Chemical Principles in the Laboratory

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Author: Emil Slowinski | Wayne C. Wolsey | Robert Rossi

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Chemical Principles in the Laboratory

Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.


Chemical Principles in the Laboratory Tenth Edition

Emil J. Slowinski Macalester College St. Paul, Minnesota

Wayne C. Wolsey Macalester College St. Paul, Minnesota

Robert C. Rossi Macalester College St. Paul, Minnesota

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Chemical Principles in the Laboratory, Tenth Edition Emil J. Slowinski, Wayne C. Wolsey, and Robert C. Rossi Acquisitions Editor: Christopher Simpson Production Manager: Matt Ballantyne Editorial Assistant: Laura Bowen

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Preface

A Few Words to the Students n spite of its many successful theories, chemistry remains, and probably always will remain, an experimental science. As the world of the computer has developed, there are those who feel that one can learn chemistry by working with software and observing reactions on the computer screen rather than in the laboratory. We do not agree with this approach. We believe that there is really no substitute for hands-on laboratory experience, if one is to learn what chemistry is all about. It is not easy to do good experimental work. It requires experience, thought, and care. As beginning students, you have not had much opportunity to do experiments. We feel that the effort you put into your laboratory sessions can pay off in many ways. You can gain a better understanding of how the chemical world works, manual dexterity in manipulating apparatus, an ability to apply mathematics to chemical systems, and, perhaps most importantly, a way of thinking that allows you to better analyze many problems in and out of science. Who knows, perhaps you will find you enjoy doing chemistry and go on to a career as a chemist, as many of our students have. In writing this manual, we attempted to illustrate many established principles of chemistry with experiments that are as interesting and challenging as possible. These principles are basic to the science, but are usually not intuitively obvious. With each experiment we introduce the theory involved, state in detail the procedures that are used, describe how to draw conclusions from your observations, and, in an Advance Study Assignment, ask you to answer questions similar to those you will encounter in the experiment. Before coming to lab, you should read over the experiment for that week, and do the Advance Study Assignment. If you prepare for lab as you should, you will get more out of it. To give an experiment a bit of a challenge, we occasionally ask you to work with chemical unknowns. In this edition, we have included a new appendix entitled “Statistical Treatment of Laboratory Data.” When professional scientists collect experimental data, especially in doing chemical analyses to determine compositions, they usually summarize their results by means of an average (the arithmetic mean) and a measure of how consistent the data are (through a statistical parameter called the standard deviation). In several experiments of this type you will be asked to calculate these quantities from your results. Your instructor may decide to make these calculations an optional part of the experiments, but these values are easily obtained on many standard calculators or via any spreadsheet (such as Excel).

I

A Few More Words, This Time to the Instructors In this, the tenth edition of our manual, we bring a new, younger co-author on board: Dr. Robert Rossi, Laboratory Supervisor for the Department of Chemistry at Macalester College. Professor Emeritus Emil Slowinski, who was deeply involved in the many earlier editions of this manual—since the inception of Chemical Principles in the Laboratory in 1969—decided not to participate in the preparation of this edition. We continue with the same format as that used in the ninth edition, but have made some important changes, some of which were suggested by users. We have changed Experiment 7, “Analysis of an Unknown Chloride,” from the familiar Mohr method (which uses potassium chromate as the indicator) to the Fajans method, which uses dichlorofluorescein as an adsorption indicator. The new method is commonly used in the industrial sector and is the official method for chloride of the Committee on Analytical Reagents of the American Chemical Society. By making this change, the experiment can be said to be “greener” in two respects—no chromate disposal problem and a lower, less costly silver nitrate concentration of 0.02 molar. See the Instructor’s Manual for some important experimental details before using this experiment. We have introduced an optional part within Experiment 14, “Heat Effects and Calorimetry,” to facilitate the experimental validation of Hess’s Law. In Experiment 29, “Synthesis and Analysis of a Coordination Compound,” we have improved the alternative procedure, A.2, for the preparation of the violet complex.


vi

Preface

In the interest of safety, throughout the manual we have substituted heptane for hexane, as heptane offers a reduced neurotoxicity hazard. As with all chemical educators, we are concerned about the use of mercury and its compounds in the general chemistry laboratory. Bismuth has now been substituted for mercury(II) in the Paper Chromatography experiment (#2) and we hope to have finalized a mercury-free means of preserving starch solutions by the time of publication of this edition. Mercury compounds continue to be used in the qualitative analysis experiments. As you will note in the student introductory section, we have increased the use of statistical analysis in several experiments. While your institution always has the option of not requiring the students to calculate means and standard deviations, we would encourage you to give your students this easy introduction to the important role of statistics in the real world. Chemical educators have frequently overlooked statistical analysis in the lower level courses and depended upon their analytical chemistry colleagues to provide this introduction. Certainly the analytical chemistry or quantitative analysis courses are where the more theoretical basis of statistics will continue to be housed, but there is no reason that the operational approach used in our Appendix VIII cannot be used by introductory chemistry students. The order of experiments in this manual makes it compatible with the order of topics in the text, Chemistry: Principles and Reactions, Seventh Edition, by William Masterton, Cecile N. Hurley, and Edward Neth. We believe, however, that the overall set of experiments should be appropriate for most standard texts in general chemistry. If this is the first time that you are using this manual, we have done what we could to make the transition to a new manual as easy as possible. The Instructor’s Manual, available in hard copy or online, contains, for each experiment, a list of required equipment and chemicals, directions for preparing solutions, and suggestions for dealing with the disposal of chemical waste. We also note the time it will take to do the experiments and an approximate cost per student. In the second part of the Instructor's Manual, directed to the lab supervisor, we offer comments and suggestions for each experiment that may be helpful, sample data and calculations, and answers to the Advance Study questions. As with any endeavor, there are many people who contribute to the effort. We gratefully acknowledge the assistance given by Amy Rice of our department, who helped develop the changes in several experiments and aided with the preparation of the manuscript for this edition. We acknowledge the assistance of Barbara Ekeberg in the preparation of the several editions of the Instructor’s Manual. We also acknowledge the input offered by those responding to a survey conducted by Joshua Duncan and the editorial offices of our publisher, Brooks-Cole. It has been a great experience being involved with this laboratory manual over its many editions. We have appreciated the support of our users. We encourage any comments, questions, or suggestions you may have. Please send them to [email protected]. Finally, we acknowledge the patient support of our families in the preparation of this edition. Emil J. Slowinski Wayne C. Wolsey Robert C. Rossi


Safety in the Laboratory

Read this section before performing any of the experiments in this manual A chemistry laboratory can be, and should be, a safe place in which to work. Yet each year in academic and industrial laboratories accidents occur that in some cases injure seriously, or even kill, chemists. Most of these accidents could have been foreseen and prevented, had the chemists involved used the proper judgment and taken proper precautions. The experiments you will be performing have been selected at least in part because they can be done safely. Instructions in the procedures should be followed carefully and in the order given. Sometimes even a change in concentration of one reagent is sufficient to change the conditions of a chemical reaction so as to make it occur in a different way, perhaps at a highly accelerated rate. So, do not deviate from the procedure given in the manual when performing experiments unless specifically told to do so by your instructor. Eye Protection: One of the simplest, and most important, things you can do in the laboratory to avoid injury

is to protect your eyes by routinely wearing safety glasses. Your instructor will tell you what eye protection to use, and you should use it. Glasses worn up on the hair may look hip, but they will not protect your eyes. If you use contact lenses, it is even more critical that you wear safety glasses as well. Chemical Reagents: Chemicals in general are toxic materials. This means that they can act as poisons or car-

cinogens (causes of cancer) if they get into your digestive or respiratory system. Never taste a chemical substance, and avoid getting any chemical on your skin. If contact should occur, wash off the affected area promptly with plenty of water. Also, wash your face and hands when you are through working in the laboratory. Never pipet by mouth; when pipetting, use a rubber bulb or other device to suck up the liquid. Avoid breathing vapors given off by reagents or reactions. If directed to smell a vapor, do so cautiously. Use a fume hood when the directions call for it. Some reagents, such as concentrated acids or bases, or bromine, are caustic, which means that they can cause chemical burns on your skin and eat through your clothing. Where such reagents are being used, we note the potential danger with a C A U T I O N : sign at that point in the procedure. Be particularly careful when carrying out that step. Always read the label on a reagent bottle before using it; there is a lot of difference between the properties of 1 M H2SO4 and those of concentrated (18 M) H2SO4. A few of the chemicals we use are flammable. These include heptane, ethanol, and acetone. Keep your Bunsen burner well away from any open beakers containing such chemicals, and be careful not to spill them on the laboratory bench where they might be easily ignited. When disposing of the chemical products from an experiment, use good judgment. Some dilute, nontoxic solutions can be poured down the sink, flushing with plenty of water. Insoluble or toxic materials should be put in the waste crocks provided for that purpose. Your lab instructor may give you instructions for treatment and disposal of the products from specific experiments. Safety Equipment: In the laboratory there are various pieces of safety equipment, which may include a safety shower, an emergency eye wash, a fire extinguisher, and a fire blanket. Learn where these items are, so that you will not have to look all over if you ever need them in a hurry. Laboratory Attire: Come to the laboratory in sensible clothing. Long, flowing robes are out, as are bare

feet. Sandals and open-toed shoes offer less protection than regular shoes. Keep long hair tied back, out of the


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Safety in the Laboratory

way of flames and reagents. Tight-fitting jewelry (especially rings) can exacerbate the harm caused by chemical exposure, and should not be worn in lab. If an Accident Occurs: During the laboratory course a few accidents will probably occur. For the most part

these will not be serious, and might involve a spilled reagent, a beaker of hot water that gets tipped over, a dropped test tube, or a small fire. A common response in such a situation is panic. A student may respond to an otherwise minor accident by doing something irrational, like running from the laboratory when the remedy for the accident is close at hand. If a mishap befalls a fellow student, watch for signs of panic and tell the student what to do; if it seems necessary, help him or her do it. Call the instructor for assistance. Most chemical spills are best handled by quickly absorbing wet material with a paper towel and then washing the area with water from the nearest sink. Use the eye wash fountain if you get something in your eye. In case of a severe chemical spill on your clothing or shoes, use the emergency shower and take off the affected clothing. In case of a fire in a beaker, on a bench, or on your clothing or that of another student, do not panic and run. Smother the fire with an extinguisher, with a blanket, or with water, as seems most appropriate at the time. If the fire is in a piece of equipment or on the lab bench and does not appear to require instant action, have your instructor put the fire out. If you cut yourself on a piece of broken glass, tell your instructor, who will assist you in treating it. A Message to Foreign Students: Many students from foreign countries take courses in chemistry before

they are completely fluent in English. If you are such a student, it may be that in some experiments you will be given directions that you do not completely understand. If that happens, do not try to do that part of the experiment by simply doing what the student next to you seems to be doing. Ask that student, or the instructor, what the confusing word or phrase means, and when you understand what you should do, go ahead. You will soon learn the language well enough, but until you feel comfortable with it, do not hesitate to ask others to help you with unfamiliar phrases and expressions.

Although we have spent considerable time here describing some of the things you should be concerned with in the laboratory from a safety point of view, this does not mean you should work in the laboratory in fear and trepidation. Chemistry is not a dangerous activity when practiced properly. Chemists as a group live longer than other professionals, in spite of their exposure to potentially dangerous chemicals. In this manual we have attempted to describe safe procedures and to employ chemicals that are safe when used properly. Many thousands of students have performed these experiments without having accidents, so you can too. However, we authors cannot be in the laboratory when you carry out the experiments to be sure that you observe the necessary precautions. You and your laboratory supervisor must, therefore, see to it that the experiments are done properly and assume responsibility for any accidents or injuries that may occur.


Contents

Experiment 1 The Densities of Liquids and Solids

1

Experiment 2 Resolution of Matter into Pure Substances, I. Paper Chromatography

7

Experiment 3 Resolution of Matter into Pure Substances, II. Fractional Crystallization 15 Experiment 4 Determination of a Chemical Formula 23 Experiment 5 Identification of a Compound by Mass Relationships

29

Experiment 6 Properties of Hydrates 35 Experiment 7 Analysis of an Unknown Chloride

41

Experiment 8 Verifying the Absolute Zero of Temperature—Determination of the Barometric Pressure 47 Experiment 9 Molar Mass of a Volatile Liquid

55

Experiment 10 Analysis of an Aluminum-Zinc Alloy Experiment 11 The Atomic Spectrum of Hydrogen

61

69

Experiment 12 The Alkaline Earths and the Halogens—Two Families in the Periodic Table

79

Experiment 13 The Geometrical Structure of Molecules—An Experiment Using Molecular Models

87


x

Contents

Experiment 14 Heat Effects and Calorimetry 97 Experiment 15 The Vapor Pressure and Heat of Vaporization of a Liquid Experiment 16 The Structure of Crystals—An Experiment Using Models Experiment 17 Classification of Chemical Substances

105

113

125

Experiment 18 Some Nonmetals and Their Compounds—Preparations and Properties Experiment 19 Molar Mass Determination by Depression of the Freezing Point Experiment 20 Rates of Chemical Reactions, I. The Iodination of Acetone Experiment 21 Rates of Chemical Reactions, II. A Clock Reaction

133

141

149

159

Experiment 22 Properties of Systems in Chemical Equilibrium—Le Châtelier’s Principle 169 Experiment 23 Determination of the Equilibrium Constant for a Chemical Reaction

181

Experiment 24 The Standardization of a Basic Solution and the Determination of the Molar Mass of an Acid

191

Experiment 25 pH Measurements—Buffers and Their Properties 199 Experiment 26 Determination of the Solubility Product of Ba(IO3)2

209

Experiment 27 Relative Stabilities of Complex Ions and Precipitates Prepared from Solutions of Copper(II) Experiment 28 Determination of the Hardness of Water

217

225


Contents

Experiment 29 Synthesis and Analysis of a Coordination Compound

231

Experiment 30 Determination of Iron by Reaction with Permanganate—A Redox Titration Experiment 31 Determination of an Equivalent Mass by Electrolysis

xi

241

247

Experiment 32 Voltaic Cell Measurements 255 Experiment 33 Preparation of Copper(I) Chloride 265 Experiment 34 Development of a Scheme for Qualitative Analysis Experiment 35 Spot Tests for Some Common Anions

277

Experiment 36 Qualitative Analysis of Group I Cations Experiment 37 Qualitative Analysis of Group II Cations Experiment 38 Qualitative Analysis of Group III Cations Experiment 39 Identification of a Pure Ionic Solid

271

285

293

301

309

Experiment 40 The Ten Test Tube Mystery 317 Experiment 41 Preparation of Aspirin 325 Experiment 42 Rate Studies on the Decomposition of Aspirin

333

Experiment 43 Analysis for Vitamin C 341


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Contents

Appendix I Vapor Pressure and Density of Liquid Water

347

Appendix II Summary of Solubility Properties of Ions and Solids

349

Appendix IIA Some Properties of the Cations in Groups I, II, and III Appendix III Table of Atomic Masses (Based on Carbon-12) Appendix IV Making Measurements—Laboratory Techniques

351

355

357

Appendix V Mathematical Considerations—Significant Figures and Making Graphs Appendix VI Suggested Locker Equipment

367

373

Appendix VII Introduction to Excel 375 Appendix VIII Statistical Treatment of Laboratory Data 379


Experiment 1 The Densities of Liquids and Solids

iven a sample of a pure liquid, we can measure many of its characteristics. Its temperature, mass, color, and volume are among the many properties we can determine. We find that, if we measure the mass and volume of different samples of the liquid, the mass and volume of each sample are related in a simple way; if we divide the mass by the volume, the result we obtain is the same for each sample, independent of its mass. That is, for samples A, B, and C, of the liquid at constant temperature and pressure,

G

MassA/VolumeA = MassB/VolumeB = MassC/VolumeC = a constant That constant, which is clearly independent of the size of the sample, is called its density, and is one of the fundamental properties of the liquid. The density of water is exactly 1.00000 g/cm3 at 4°C, and is slightly less than one at room temperature (0.9970 g/cm3 at 25°C). Densities of liquids and solids range from values that are less than that of water to values that are much greater. Osmium metal has a density of 22.5 g/cm3 and is probably the densest material known at ordinary pressures. In any density determination, two quantities must be determined—the mass and the volume of a given quantity of matter. The mass can easily be determined by weighing a sample of the substance on a balance. The quantity we usually think of as “weight” is really the mass of a substance. In the process of “weighing” we find the mass, taken from a standard set of masses, that experiences the same gravitational force as that experienced by the given quantity of matter we are weighing. The mass of a sample of liquid in a container can be found by taking the difference between the mass of the container plus the liquid and the mass of the empty container. The volume of a liquid can easily be determined by means of a calibrated container. In the laboratory a graduated cylinder is often used for routine measurements of volume. Accurate measurement of liquid volume is made using a pycnometer, which is simply a container having a precisely definable volume. The volume of a solid can be determined by direct measurement if the solid has a regular geometric shape. Such is not usually the case, however, with ordinary solid samples. A convenient way to determine the volume of a solid is to measure accurately the volume of liquid displaced when an amount of the solid is immersed in the liquid. The volume of the solid will equal the volume of liquid which it displaces. In this experiment we will determine the density of a liquid and a solid by the procedure we have outlined. First we weigh an empty flask and its stopper. We then fill the flask completely with water, measuring the mass of the filled stoppered flask. From the difference in these two masses we find the mass of water and then, from the known density of water, we determine the volume of the flask. We empty and dry the flask, fill it with an unknown liquid, and weigh again. From the mass of the liquid and the volume of the flask we find the density of the liquid. To determine the density of an unknown solid metal, we add the metal to the dry empty flask and weigh. This allows us to find the mass of the metal. We then fill the flask with water, leaving the metal in the flask, and weigh again. The increase in mass is that of the added water; from that increase, and the density of water, we calculate the volume of water we added. The volume of the metal must equal the volume of the flask minus the volume of water. From the mass and volume of the metal we calculate its density. The calculations involved are outlined in detail in the Advance Study Assignment.


2

Experiment 1

The Densities of Liquids and Solids

Experimental Procedure A. Mass of a Coin After you have been shown how to operate the analytical balances in your laboratory, read the section on balances in Appendix IV. Take a coin and measure its mass to 0.0001 g. Record the mass on the Data page. If your balance has a TARE bar, use it to re-zero the balance. Take another coin and weigh it, recording its mass. Remove both coins, zero the balance, and weigh both coins together, recording the total mass. If you have no TARE bar on your balance, add the second coin and measure and record the mass of the two coins. Then remove both coins and find the mass of the second one by itself. When you are satisfied that your results are those you would expect, go to the stockroom and obtain a glass-stoppered flask, which will serve as a pycnometer, and samples of an unknown liquid and an unknown metal.

B. Density of a Liquid If your flask is not clean and dry, clean it with detergent solution and water, rinse it with a few cubic centimeters of acetone, and dry it by letting it stand for a few minutes in the air or by gently blowing compressed air into it for a few moments. Weigh the dry flask with its stopper on the analytical balance, or the top-loading balance if so directed, to the nearest milligram. Fill the flask with distilled water until the liquid level is nearly to the top of the ground surface in the neck. Put the stopper in the flask in order to drive out all the air and any excess water. Work the stopper gently into the flask, so that it is firmly seated in position. Wipe any water from the outside of the flask with a towel and soak up all excess water from around the top of the stopper. Again weigh the flask, which should be completely dry on the outside and full of water, to the nearest milligram. Given the density of water at the temperature of the laboratory and the mass of water in the flask, you should be able to determine the volume of the flask very precisely. Empty the flask, dry it, and fill it with your unknown liquid. Stopper and dry the flask as you did when working with the water, and then weigh the stoppered flask full of the unknown liquid, making sure its surface is dry. This measurement, used in conjunction with those you made previously, will allow you to accurately determine the density of your unknown liquid.

C. Density of a Solid Pour your sample of liquid from the flask into its container. Rinse the flask with a small amount of acetone and dry it thoroughly. Add small chunks of the metal sample to the flask until the flask is at least half full. Weigh the flask, with its stopper and the metal, to the nearest milligram. You should have at least 50 g of metal in the flask. Leaving the metal in the flask, fill the flask with water and then replace the stopper. Roll the metal around in the flask to make sure that no air remains between the metal pieces. Refill the flask if necessary, and then weigh the dry, stoppered flask full of water plus the metal sample. Properly done, the measurements you have made in this experiment will allow a calculation of the density of your metal sample that will be accurate to about 0.1%. Pour the water from the flask. Put the metal in its container. Dry the flask and return it with its stopper and your metal sample, along with the sample of unknown liquid, to the stockroom.

DISPOSAL OF REACTION PRODUCTS.


Name ____________________________________

Section _________________________________

Experiment 1

Data and Calculations: The Densities of Liquids and Solids A. Mass of coin 1 ____________ g

Mass of coin 2 ____________ g

Mass of coins 1 and 2 weighed together ____________ g What general law is illustrated by the results of this experiment?

B. Density of unknown liquid Mass of empty flask plus stopper

____________ g

Mass of stoppered flask plus water

____________ g

Mass of stoppered flask plus liquid

____________ g

Mass of water

____________ g

Temperature in the laboratory

____________ °C

Volume of flask (obtain density of water from Appendix I)

____________ cm3

Mass of liquid

____________ g

Density of liquid

____________ g/cm3

To how many significant figures can the liquid density be properly reported? (See Appendix V.)

____________

C. Density of unknown metal Mass of stoppered flask plus metal

____________ g

Mass of stoppered flask plus metal plus water

____________ g

Mass of metal

____________ g

Mass of water

____________ g

Volume of water

____________ cm3 (continued on following page)


4

Experiment 1


Volume of metal

____________ cm3

Density of metal

____________ g/cm3

To how many significant figures can the density of the metal be properly reported?

____________

Explain why the value obtained for the density of the metal is likely to have a larger percentage error than that found for the liquid.

Unknown liquid no. ____________

Unknown solid no. ____________


Name ____________________________________

Section _________________________________

Experiment 1

Advance Study Assignment: Densities of Solids and Liquids The advance study assignments in this laboratory manual are designed to assist you in making the calculations required in the experiment you will be doing. We do this by furnishing you with sample data and showing in some detail how that data can be used to obtain the desired results. In the advance study assignments we will often include the guiding principles as well as the specific relationships to be employed. If you work through the steps in each calculation by yourself, you should have no difficulty when you are called upon to make the necessary calculations on the basis of the data you obtain in the laboratory.

1. Finding the volume of a flask. A student obtained a clean, dry glass-stoppered flask. She weighed the flask and stopper on an analytical balance and found the total mass to be 33.695 g. She then filled the flask with water and obtained a mass for the full stoppered flask of 64.356 g. From these data, and the fact that at the temperature of the laboratory the density of water was 0.9978 g/cm3, find the volume of the stoppered flask. a.

First we need to obtain the mass of the water in the flask. This is found by recognizing that the mass of a sample is equal to the sum of the masses of its parts. For the filled stoppered flask: Mass of filled stoppered flask = mass of empty stoppered flask + mass of water, so mass of water = mass of filled flask – mass of empty flask Mass of water = ____________ g – ____________ g = ____________ g Many mass and volume measurements in chemistry are made by the method used in 1a. This method is called measuring by difference, and is a very useful one.

b. The density of a pure substance is equal to its mass divided by its volume: mass Density = volume

mass or volume = density

The volume of the flask is equal to the volume of the water it contains. Since we know the mass and density of the water, we can find its volume and that of the flask. Make the necessary calculation.

Volume of water = volume of flask = ____________ cm3

2. Finding the density of an unknown liquid. Having obtained the volume of the flask, the student emptied the flask, dried it, and filled it with an unknown whose density she wished to determine. The mass of the stoppered flask when completely filled with liquid was 58.540 g. Find the density of the liquid. a.

First we need to find the mass of the liquid by measuring by difference: Mass of liquid = ____________ g – ____________ g = ____________ g (continued on following page)


6

Experiment 1


b. Since the volume of the liquid equals that of the flask, we know both the mass and volume of the liquid and can easily find its density using the equation in 1b. Make the calculation.

Density of liquid = ____________ g/cm3

3. Finding the density of a solid. The student then emptied the flask and dried it once again. To the empty flask she added pieces of a metal until the flask was about three-fourths full. She weighed the stoppered flask and its metal contents and found that the mass was 144.076 g. She then filled the flask with water, stoppered it, and obtained a total mass of 162.292 g for the flask, stopper, metal, and water. Find the density of the metal. a. To find the density of the metal we need to know its mass and volume. We can easily obtain its mass by the method of differences: Mass of metal = ____________ g − ____________ g = ____________ g b. To determine the volume of metal, we note that the volume of the flask must equal the volume of the metal plus the volume of water in the filled flask containing both metal and water. If we can find the volume of water, we can obtain the volume of metal by the method of differences. To obtain the volume of the water we first calculate its mass: Mass of water = mass of (flask + stopper + metal + water) − mass of (flask + stopper + metal) Mass of water = ____________ g − ____________ g = ____________ g The volume of water is found from its density, as in 1b. Make the calculation.

Volume of water = ____________ cm3 c. From the volume of the water, we calculate the volume of metal:

Volume of metal = volume of flask − volume of water Volume of metal = ____________ cm3 − ____________ cm3 = ____________ cm3 From the mass of and volume of metal, we find the density, using the equation in 1b. Make the calculation.

Density of metal = ____________ g/cm3 Now go back to Question 1 and check to see that you have reported the proper number of significant figures in each of the results you calculated in this assignment. Use the rules on significant figures as given in your chemistry text or Appendix V.


Experiment 2 Resolution of Matter into Pure Substances, I. Paper Chromatography

he fact that different substances have different solubilities in a given solvent can be used in several ways to effect a separation of substances from mixtures in which they are present. We will see in an upcoming experiment how fractional crystallization allows us to obtain pure substances by relatively simple procedures based on solubility properties. Another widely used resolution technique, which also depends on solubility differences, is chromatography. In the chromatographic experiment a mixture is deposited on some solid adsorbing substance, which might consist of a strip of filter paper, a thin layer of silica gel on a piece of glass, some finely divided charcoal packed loosely in a glass tube, or even a resin coating microscopic glass beads which are then packed into a length of copper tubing. The components of a mixture are adsorbed on the solid to varying degrees, depending on the nature of the component, the nature of the adsorbent, and the temperature. A solvent is then made to flow through the adsorbent solid under applied or gravitational pressure or by the capillary effect. As the solvent passes the deposited sample, the various components tend, to varying extents, to be dissolved and swept along the solid. The rate at which a component will move along the solid depends on its relative tendency to be dissolved in the solvent and adsorbed on the solid. The net effect is that, as the solvent passes slowly through the solid, the components separate from each other and move along as rather diffuse zones. With the proper choice of solvent and adsorbent, it is possible to resolve many complex mixtures by this procedure. If necessary, we can usually recover a given component by identifying the position of the zone containing the component, removing that part of the solid from the system, and eluting the desired component with a suitable (good) solvent. The name given to a particular kind of chromatography depends upon the manner in which the experiment is conducted. Thus, we have column, thin-layer, paper, and gas chromatography, all in very common use (Fig. 2.1). Chromatography in its many possible variations offers the chemist one of the best methods, if not the best method, for resolving a mixture into pure substances, regardless of whether that mixture consists of a gas, a volatile liquid, or a group of nonvolatile, relatively unstable, complex organic compounds.

T

Figure 2.1 This is a gas chromatogram of a sample of unleaded gasoline. Each peak corresponds to a different molecule, so gasoline has many components, at least 50, each of which can be identified. The molar masses vary from about 50 to about 150, with the largest peak, at about 3 min, being due to toluene, C6H5CH3. Sample size for the chromatogram was less than 10–6 grams, 0 , since Δt = tfinal − tinitial.) Given the specific heat of water, we can find the positive 2 specific heat of the metal by Equation 3. We will use this procedure to obtain the specific heat of an unknown metal. The specific heat of a metal is related in a simple way to its molar mass. Dulong and Petit discovered many years ago that about 25 joules were required to raise the temperature of one mole of many metals by 1°C. This relation, shown in Equation 4, is known as the Law of Dulong and Petit: MM ≅

25 S. H.( J/g°C)

(4)

where MM is the molar mass of the metal. Once the specific heat of the metal is known, the approximate molar mass can be calculated by Equation 4. The Law of Dulong and Petit was one of the few rules available to early chemists in their studies of molar masses.

B. Heat of Reaction When a chemical reaction occurs in aqueous solution, the situation is similar to that which is present when a hot metal sample is put into water. With such a reaction there is an exchange of heat between the reaction mixture and the solvent, water. As in the specific heat experiment, the heat flow for the reaction mixture is equal in magnitude but opposite in sign to that for the water. The heat flow associated with the reaction mixture is also equal to the enthalpy change, ΔH, for the reaction, so we obtain the equation qreaction = ΔHreaction = −qH O 2

(5)

By measuring the mass of the water used as solvent, and by observing the temperature change that the water undergoes, we can find qH2O by Equation 1 and ΔH by Equation 5. If the temperature of the water goes up, heat has been given off by the reaction mixture, so the reaction is exothermic; qH2O is positive and ΔH is negative. If the temperature of the water goes down, the reaction mixture has absorbed heat from the water and the reaction is endothermic. In this case qH2O is negative and ΔH is positive. Both exothermic and endothermic reactions are observed. One of the simplest reactions that can be studied in solution occurs when a solid is dissolved in water. As an example of such a reaction note the solution of NaOH in water: NaOH(s) → Na+(aq) + OH−(aq); ΔH = ΔHsolution

(6)

When this reaction occurs, the temperature of the solution becomes much higher than that of the NaOH and water that were used. If we dissolve a known amount of NaOH in a measured amount of water in a calorimeter, and measure the temperature change that occurs, we can use Equation 1 to find qH2O for the reaction and use Equation 5 to obtain ΔH. Noting that ΔH is directly proportional to the amount of NaOH used, we can easily calculate ΔHsolution for either a gram or a mole of NaOH. In the second part of this experiment you will measure ΔHsolution for an unknown ionic solid. Chemical reactions often occur when solutions are mixed. A precipitate may form, in a reaction opposite in direction to that in Equation 6. A very common reaction is that of neutralization, which occurs when an acidic solution is mixed with one that is basic. In the last part of this experiment you will measure the heat effect when a solution of HCl, hydrochloric acid, is mixed with one containing NaOH, sodium hydroxide, which is basic. The heat effect is quite large, and is the result of the reaction between H+ ions in the HCl solution with OH− ions in the NaOH solution: H+(aq) + OH− (aq) → H2O

ΔH = ΔHneutralization

(7)


Experiment 14

Heat Effects and Calorimetry

99

Experimental Procedure A. Specific Heat From the stockroom obtain a calorimeter, a sensitive thermometer, a sample of metal in a large stoppered test tube, and a sample of unknown solid. (The thermometer is expensive, so be careful when handling it.) The calorimeter consists of two nested expanded polystyrene coffee cups fitted with a Styrofoam cover. There are two holes in the cover: one for a thermometer and the other for a glass stirring rod with a perpendicular loop on one end. Assemble the experimental setup as shown in Figure 14.1. Fill a 400-cm3 beaker two thirds full of water and begin heating it to boiling. While the water is heating, weigh your sample of unknown metal in the large stoppered test tube to the nearest 0.1 g on a top-loading or triple-beam balance. Pour the metal into a dry container and weigh the empty test tube and stopper. Replace the metal in the test tube and put the loosely stoppered tube into the hot water in the beaker. The water level in the beaker should be high enough that the top of the metal is below the water surface. Continue heating the metal in the water for at least 10 minutes after the water begins to boil, to ensure that the metal attains the temperature of the boiling water. Add small amounts of water as necessary to maintain the water level. While the water is boiling, weigh the calorimeter to 0.1 g. Place about 40 cm3 of water in the calorimeter and weigh again. Insert the stirrer and thermometer into the cover and put it on the calorimeter. If a glass thermometer is being used, the thermometer bulb should be completely immersed. Measure the temperature of the water in the calorimeter to 0.1°C. Take the test tube out of the beaker of boiling water, remove the stopper, and pour the metal into the water in the calorimeter. Be careful that no water adhering to the outside of the test tube runs into the calorimeter when you are pouring the metal. Replace the calorimeter cover and agitate the water as best you can with the glass stirrer. Record to 0.1°C the maximum temperature reached by the water. Repeat the experiment, using about 50 cm3 of water in the calorimeter. Be sure to dry your metal before reusing it; this can be done by heating the metal briefly in the test tube in boiling water and then pouring the metal onto a paper towel to drain. You can dry the hot test tube with a little compressed air.

Figure 14.1 Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

100

Experiment 14


B. Heat of Solution Place about 50 cm3 of distilled water in the calorimeter and weigh as in the previous procedure. Measure the temperature of the water to 0.1°C. The temperature should be within a degree or two of room temperature. In a small beaker weigh out about 5 g of the solid compound assigned to you. Make the weighing of the beaker and of the beaker plus solid to 0.1 g. Add the compound to the calorimeter. Stirring continuously and occasionally swirling the calorimeter, determine to 0.1°C the maximum or minimum temperature reached as the solid dissolves. Check to make sure that all the solid is dissolved. A temperature change of at least 5 degrees should be obtained in this experiment. If necessary, repeat the experiment, increasing the amount of solid used. DISPOSAL OF REACTION PRODUCTS.

Dispose of the solution in Part B as directed by your instructor.

C. Heat of Neutralization Rinse out your calorimeter with distilled water, pouring the rinse into the sink. In a graduated cylinder, measure out 25.0 cm3 of 2.00 M HCl; pour that solution into the calorimeter. Rinse out the cylinder with distilled water, and measure out 25.0 cm3 of 2.00 M NaOH; pour that solution into a dry 50-mL beaker. Measure the temperature of the acid and of the base to ±0.1°C, making sure to rinse and dry your thermometer before immersing it in the solutions. Put the thermometer back in the calorimeter cover. Pour the NaOH solution into the HCl solution and put on the cover of the calorimeter. Stir the reaction mixture, and record the maximum temperature that is reached by the neutralized solution. Calculate the heat of neutralization. It is necessary to slightly modify Equation 1 when calculating qH O. Use a value of 1.02 g/mL as an average density for all solutions in this experiment; thus the mass of the solution is obtained by multiplying 50.0 mL by 1.02 g/mL. We will continue to use 4.18 J/gºC as the specific heat value. 2

D. Hess’ Law

Optional

Using the same procedure as in Part C, measure the heat of neutralization of 2.00 M acetic acid (HAc). You should use 25.0 mL of the acetic acid solution in place of the 2.00 M HCl used in Part C. Calculate the molar heat of solution of acetic acid, HC2H3O2, again using 1.02 g/mL as the average density for all solutions. Record your data and calculations on a separate sheet of paper. Write out the net ionic equation for the strong acid-strong base neutralization and the net ionic equation for the acetic acid (weak acid; use HAc to represent undissociated acetic acid)-strong base neutralization reaction. In each case, show the value of the measured molar heat of reaction on the right side of the equation. Remember that the heat flow for an exothermic reaction is negative. Combine (via addition and cancellation of any species that appears on opposite sides of the final equation) the two net ionic equations in such a way as to end up with an equation representing the dissociation reaction of the weak acid HAc to yield H+(aq) and Ac−(aq). If you needed to reverse one of the equations, the heat flow value will change signs. As you add the two equations you should add the two heat flow values, to give the final value for the desired reaction. Just as the dissociation of water to form H+(aq) and OH−(aq) is expected to be endothermic, so should the corresponding dissociation reaction for a weak acid, such as acetic acid. If you did the above exercise correctly, you have just illustrated HESS’S LAW. This law states that the value of ΔH for a reaction is the same whether it occurs directly, or in a series of steps. Thus, if a thermochemical equation can be expressed as the sum of two equations, Equation (3) = Equation (1) + Equation (2), then ΔH3 = ΔH1 + ΔH2. Optional Experiment: Measure the heat of neutralization of household vinegar in its reaction with NaOH solution. Assuming that vinegar is about 5 mass percent acetic acid, find the molar heat of neutralization of that acid. When you have completed the experiment, you may pour the neutralized solution down the sink. Rinse the calorimeter and thermometer with water, and return them, along with the metal sample to the stockroom.


Name ____________________________________

Section _________________________________

Experiment 14

Data and Calculations: Heat Effects and Calorimetry

A. Specific Heat Trial 1

Trial 2

Mass of stoppered test tube plus metal

____________ g

→

____________ g

Mass of test tube and stopper

____________ g

→

____________ g

Mass of calorimeter

____________ g

→

____________ g

Mass of calorimeter and water

____________ g

____________ g

Mass of water

____________ g

____________ g

Mass of metal

____________ g

Initial temperature of water in calorimeter

____________ °C

Initial temperature of metal (assume 100°C unless directed to do otherwise)

____________ °C

Equilibrium temperature of metal and water in calorimeter

____________ °C

____________ °C

Δtwater (tfinal − tinitial)

____________ °C

____________ °C

Δtmetal

____________ °C

____________ °C

qH2O

____________ J

____________ J

Specific heat of the metal (Eq. 3)

____________ J/g°C

____________ J/g°C

Approximate molar mass of metal

____________

____________

Unknown no.

→

____________ g ____________ °C

→

____________ °C

____________

B. Heat of Solution Mass of calorimeter plus water

____________ g

Mass of beaker

____________ g (continued on following page)


102

Experiment 14


Mass of beaker plus solid

____________ g

Mass of water, mH2O

____________ g

Mass of solid, ms

____________ g

Original temperature

____________ °C

Final temperature

____________ °C

qH2O for the reaction (Eq. 1) (S.H. = 4.18 J/g°C)

____________ joules

ΔH for the reaction (Eq. 5)

____________ joules

The quantity you have just calculated is approximately* equal to the heat of solution of your sample. Calculate the heat of solution per gram of solid sample. ΔHsolution = ____________ joules/g The solution reaction is endothermic exothermic. (Underline correct answer.) Give your reasoning. Solid unknown no. ____________ Optional Formula of compound used (if furnished) ____________ Molar mass ____________ g

Heat of solution per mole of compound

____________ kJ

C. Heat of Neutralization Original temperature of HCl solution

____________ °C

Original temperature of NaOH solution

____________ °C

Final temperature of neutralized mixture

____________ °C

Change in temperature. Δt (take average of the original temperatures of HCl and NaOH)

____________ °C

qH2O (assume 50.0 mL of solution and use average density of 1.02 g/mL)

____________ J

ΔH for the neutralization reaction

____________ J

ΔH per mole of H+ and OH− ions reacting

____________ kJ

*The value of ΔH will be approximate for several reasons. One is that we do not include the amount of heat absorbed by the solute. This effect is smaller than the likely experimental error, and thus we will ignore it.


Name ____________________________________

Section _________________________________

Experiment 14

Advance Study Assignment: Heat Effects and Calorimetry 1. A metal sample weighing 124.10 g and at a temperature of 99.3°C was placed in 42.87 g of water in a calorimeter at 23.9°C. At equilibrium the temperature of the water and metal was 41.6°C. a. What was Δt for the water? (Δt = tfinal − tinitial) ____________ °C b. What was Δt for the metal? ____________ °C

c.

How much heat flowed into the water? ____________ joules

d. Taking the specific heat of water to be 4.18 J/g°C, calculate the specific heat of the metal, using Equation 3. ____________ joules/g°C

e. What is the approximate molar mass of the metal? (Use Eq. 4.) ____________ g

2. When 5.12 g of NaOH were dissolved in 51.55 g water in a calorimeter at 24.5°C, the temperature of the solution went up to 49.8°C. a. Is this solution reaction exothermic? ____________ Why?

b. Calculate qH2O , using Equation 1. ____________ joules c. Find ΔH for the reaction as it occurred in the calorimeter (Eq. 5). ΔH = ____________ joules

(continued on following page)


104

Experiment 14


d. Find ΔH for the solution of 1.00 g NaOH in water. ΔH = ____________ joules/g

e.

Find ΔH for the solution of 1 mole NaOH in water. ΔH = ____________ joules/mole

f.

Given that NaOH exists as Na+ and OH− ions in solution, write the equation for the reaction that occurs when NaOH is dissolved in water.

g. Given the following heats of formation, ΔHf , in kJ per mole, as obtained from a table of ΔHf data, calculate ΔH for the reaction in Part f. Compare your answer with the result you obtained in Part e. NaOH(s), −425.6; Na+(aq), −240.1; OH−(aq), −230.0 ΔH = ____________ kJ


Experiment 15 The Vapor Pressure and Heat of Vaporization of a Liquid*

f we pour a liquid into an open container, which we then close, we find that some of the liquid will evaporate into the air in the container. The vapor will after a short time establish a partial pressure that holds constant as long as the temperature does not change. That pressure is called the vapor pressure of the liquid. It does not change if there are no other gases present. In the air we breathe, there is some water vapor, usually with a partial pressure well below the vapor pressure. The relative humidity is equal to 100% times the partial pressure of water vapor divided by the vapor pressure at that temperature. On a hot, sticky day the relative humidity is high, perhaps 80 or 90%, and the air is nearly saturated with water vapor; we feel uncomfortable. If we raise the temperature of a liquid, its vapor pressure increases, ever more rapidly. At the normal boiling point of the liquid, the vapor pressure becomes one atmosphere. That would be the pressure at 100°C in a closed container in which there is just water and its vapor and no air. At high temperatures, the vapor pressure can become very large, reaching about 5 atm for water at 150°C and 15 atm at 200°C. A liquid that boils at a low temperature, say −50°C, will have a very large vapor pressure at room temperature, and confining it at 25°C in a closed container may well cause an explosion. Failure to recognize this fact has caused several scientists of the authors’ acquaintance to have serious accidents. There is an equation relating the vapor pressure of a liquid to the Kelvin temperature. It is called the Clapeyron Equation and has the form:

I

ln P = − ΔH/RT + a constant

(1)

If we plot ln P vs the reciprocal of the Kelvin temperature we obtain a straight line, the slope of which equals −ΔH/R. In the equation ln P is the natural logarithm of P, and the units of ΔH and R must match. P may be in atm, mm Hg, bars, or any other pressure unit. There are many methods for measuring vapor pressure, most of which are not intuitively obvious. In this experiment we will use one of the simplest, and yet most accurate, methods. We will inject a measured sample of dry air into a graduated pipet containing the liquid we are studying. We measure the volume of the air-vapor bubble that forms, which will be larger than that of the injected air, since some of the liquid vaporizes. If we raise the temperature the volume will go up as more and more liquid vaporizes. The total pressure in the bubble remains equal to the barometric pressure, so if we can calculate the partial pressure of air in the bubble, we can find the vapor pressure of the liquid at that temperature by simple subtraction. (There is no danger of an explosion in this experiment, in case you are worried.)

Experimental Procedure You may work in pairs on this experiment. Both members of the pair must submit a report. From the stockroom obtain a modified 1-mL graduated pipet, a Pasteur pipet, a digital thermometer, and a sample of a liquid unknown, whose density will be given to you.

*The general method used in this experiment is one described by DeMuro, Margarian, Mkhikian, No, and Peterson, J Chem Ed 76:1113–1116, 1999.


106

Experiment 15

The Vapor Pressure and Heat of Vaporization of a Liquid

Setting the Stage The modified graduated pipet is the instrument you will use; let’s call it the device. Weigh the empty dry device accurately on the analytical balance, and record its mass. Using the Pasteur pipet, carefully fill the device completely with your liquid unknown. Dry off the outside and reweigh; record that mass. Record all your data in this experiment on the blank Data and Calculations page, noting in each case what it was you measured. Underline all of the quantities you actually measured. Using a syringe, your instructor will inject 0.200 mL of dry air at the closed end of the inverted device, forming a bubble. If, after the syringe is removed, the device is not full at the open end, add unknown with your Pasteur pipet until it is full. Then carefully dry and reweigh the device, and once again record the mass. Place the device, bubble up, in a vertical position, while holding it near the open end; wait until the liquid level becomes steady. Read and record the level of the liquid meniscus in the bubble as accurately as possible (this is a crucial step, so take your time, and try to get the reading to the nearest 0.002 mL; note that the numbers on the pipet are in tenths of milliliters, and the distance between graduations is 0.01 mL). Record the barometric pressure in mm Hg and the temperature in the lab in °C.

Finding the Vapor Pressure of the Liquid at Room Temperature From the data you have obtained so far you can calculate the vapor pressure of your unknown at the temperature in the lab. Using the mass measurements you can find the mass of the liquid in the full pipet and the mass of liquid driven out. The volume of the bubble is equal to the mass of liquid driven out divided by its density. This volume is larger than that of the air injected, since some of the liquid vaporizes into the bubble. The partial pressure of the air in the bubble is equal, by Boyle’s Law, to the barometric pressure times the initial volume of the air divided by the volume of the bubble. The vapor pressure of the liquid is obtained by subtracting the partial pressure of the air from the barometric pressure. Record all your results. You can make these calculations while performing the rest of the experiment. There are several steps in finding the vapor pressure, so make sure you know what you are doing before making a calculation.

Finding the Vapor Pressure at Other Temperatures We will use a water bath to study the vapor pressure as a function of temperature. Set up the bath as shown in Figure 15.1. Use a 1000-mL beaker that you can heat with a Bunsen burner or a hot plate. Fill the beaker to near the top with warm water from the tap (between 40° and 45°C). Fill a regular (18 × 150 mm) test tube with your unknown liquid to within 2 cm of the top, and gently drop your device, bubble up, into the tube, first making sure that no air bubble remains at the open end. Clamp the test tube as shown in Figure 15.1, so that the water level in the bath is as high as the liquid level in the test tube. Insert the digital thermometer in the test tube, holding it in position with a loosely-fitting stopper. Lay your lab thermometer in the water in the beaker. If you are heating your bath with a burner, put a stirring rod in the water. With a hot plate use a magnetic stirrer. The temperature of the water bath should be about 40°C for the first measurement. Heat or cool the bath if necessary, with stirring, to get to within a few degrees of that value. It should stabilize within a few minutes. When the temperature on the digital thermometer holds steady for at least a minute, record the temperature. Note how closely the temperature matches that of the water bath; they should be close to one another when steady. Read the meniscus level in the graduated pipet as carefully as you can. Record the level and the temperature. Heat the water in the large beaker, with stirring. When the total volume of the bubble has gone up from its last value by about 0.05 mL, as noted by the change in meniscus levels, stop heating. (The temperature at that point will be roughly 10°C above the first value.) The temperature will continue to rise for a few minutes as you stir, but will finally level off. The temperature on the digital thermometer and the bath thermometer should be about equal, and remain constant for at least a minute. Then read the level of the meniscus as accurately as you can, and record its value and the temperature.


Experiment 15


107

Digital thermometer Loose-fitting stopper Clamp

1000-mL beaker Bubble Water Unknown liquid Device

Magnetic stirr bar

Hot plate and magnetic stirrer

Figure 15.1

Repeat the measurements you have just made at two higher temperatures, at points where the volume has increased by another 0.05 mL, and then by 0.10 mL more. Each time stop heating at that point, let the temperature level off, and when both temperature and bubble volume are steady for at least a minute, record the meniscus level and the temperature. Do not, do not, do not heat the sample to the point that the bubble volume exceeds the volume of the device. If that appears to be likely to happen, add some cold water to the bath and stir. When things have calmed down, proceed to make a measurement. The data you have obtained will allow you to calculate the vapor pressure of the liquid at each temperature you used. The method is as noted earlier. First find the volume that the air sample would occupy at each temperature and barometric pressure (use Charles’ Law). The partial pressure of air in the bubble will equal barometric pressure times the ratio of the volume you just calculated to the total bubble volume, which you find by noting how much the volume increased over the value at lab temperature (the difference between meniscus levels at the two temperatures). Calculate and record the partial pressures of air and vapor pressures as you perform the next experiment.

Direct Measurement of the Boiling Point In the last part of this experiment you will measure the boiling point of the liquid at the barometric pressure. Pour the liquid sample in the regular test tube into a large test tube. Determine the boiling point of the liquid using the apparatus shown in Figure 15.2. The thermometer bulb should be just above the liquid surface. Heat the water in the bath until the liquid in the tube boils gently, with the vapor condensing at least 5 cm below the top of the test tube. Boiling chips may help in keeping the liquid boiling smoothly. As the boiling proceeds there will be some condensation on the thermometer and droplets will be falling from the thermometer bulb. After a minute or two the temperature should become reasonably steady. Record the temperature. The liquid you are using may be flammable and toxic, so you should not inhale the vapor unnecessarily. Do not heat the water bath so strongly that condensation of vapors occurs only at the top of the test tube.


108

Experiment 15


Split stopper Thermometer

Condensing vapors

Water bath Boiling chips Iron ring and iron gauze

Bunsen burner

Figure 15.2

Finding the Heat of Vaporization of the Liquid When you have completed the calculations described in the previous sections, you should have a list of the vapor pressures of the liquid at each of the temperatures you used. To find the molar heat of vaporization of the liquid we use Equation 1. First make a table listing ln P (the natural logarithm of P), the temperature T in Kelvin, and 1/T as found from your data. Make a graph of ln P vs 1/T, using the graph paper provided or software such as Excel. The slope of the line, which should be straight, is equal to −ΔH/R, the molar heat of vaporization divided by the gas constant R. If you want to express ΔH in joules/mole, R will equal 8.314 J/mole K. Compare the value of the boiling point as measured directly with that you can calculate from your graph. The value of 1/T at the boiling point will occur where ln P equals ln Pbarometric. Finally, make a graph of the vapor pressure as a function of the temperature in °C. Use Excel or the graph paper provided. Connect the points with a smooth curve. Comment on how VP varies with t. Include your graphs with your report. When you are finished with the experiment, pour the liquid from the test tube back into its container and return it to the stockroom along with the digital thermometer, the device, and the Pasteur pipet.


Name ____________________________________

Section _________________________________

Experiment 15

Data and Calculations: Vapor Pressure and Heat of Vaporization For this experiment you will be responsible for recording and processing your data. For each entry note the units and the quantity that was measured, along with temperature and pressure if necessary. Lay the page out so that you can find needed items easily. Put all the data at the top of the page, leaving the bottom part for calculations. Show the equations you used to determine calculated values, first with symbols and then with observed data. Make a table of the observed vapor pressures and temperatures, and the values of ln P and 1/T. Report the value of ΔH as calculated, and the boiling point as found directly and as calculated from the graph you prepared. Use the back of this page if necessary. Include your graph in your report. Excel can be used for all calculations and graphs. Your report can be the spreadsheet.


Name ____________________________________

Section _________________________________

Experiment 15

Vapor Pressure and Heat of Vaporization of Liquids (Data and Calculations)


Name ____________________________________

Section _________________________________

Experiment 15

Advance Study Assignment: Vapor Pressure and Heat of Vaporization of Liquids 1. In an experiment to measure the vapor pressure of methanol, the following data were obtained: Mass of empty device Mass full of methanol Mass after adding 0.200 mL air Bubble meniscus reading

2.6460 g 3.2035 g 3.0170 g 0.132 mL

Pbar = 749 mm Hg t = 23.0°C density of methanol = 0.790 g/mL

How many grams of methanol were in the full device?

____________ g

How many grams were driven out by the air?

____________ g

What is the volume of the methanol driven out?

____________ mL

What is the volume of the bubble?

____________ mL

Find the partial pressure of air in the bubble. (Note that the mass of air remains the same, but it occupies a larger volume than 0.200 mL since some vapor entered the bubble. Total P is 749 mm Hg.) ____________ mm Hg What is the partial pressure of vapor in the bubble? (This is the vapor pressure of methanol at 23.0°C!) ____________ mm Hg 2. When the device is heated to 42.5°C, the meniscus reading increases to 0.250 mL. a. What is the new volume of the bubble?

____________ mL

b. Why does the volume increase? (Two reasons)

c. What is the vapor pressure of ethanol at 42.5°C? ____________ mm Hg



Experiment 16 The Structure of Crystals— An Experiment Using Models

f one examines the crystals of an ordinary substance like table salt, using a magnifying glass or a microscope, one finds many cubic particles, among others, in which planes at right angles are present. This is the situation with many common solids. The regularity we see implies a deeper regularity in the arrangement of atoms or ions in the solid. Indeed, when we study crystals by X-ray diffraction, we find that the atomic nuclei are present in remarkably symmetrical arrays, which continue in three dimensions for thousands or millions of units. Substances having a regular arrangement of atom-size particles in the solid are called crystalline, and the solid material consists of crystals. This experiment deals with some of the simpler arrays in which atoms or ions occur in crystals, and what these arrays can tell us about such properties as atomic sizes, densities of solids, and the efficiency of packing of particles. Many crystals are complex, almost beyond belief. We will limit ourselves to the simplest crystals, those which have cubic structures. They are by far the easiest to understand, yet they exhibit many of the interesting properties of more complicated structures. We will further limit our discussion to substances containing only one or two kinds of atoms, so we will be working with the crystals of some of the elements and some binary compounds. The atoms in crystals occur in planes. Sometimes a crystal will cleave readily at such planes. This is the reason the cubic structure of salt is so apparent. Let us begin our study of crystals with a simple example, a two-dimensional crystal in which all of the atoms lie in a square array in a plane, in the manner shown below:

I

The first thing that you need to realize is that the array goes on essentially forever, in both directions. In this array is always the same kind of atom, so all the atoms are equivalent. The distance, d, between atoms in the vertical direction is the same as in the horizontal, so if you knew either distance you could generate the array completely. In crystal studies we select a small section of the array, called the unit cell, that represents the array in the sense that by moving the unit cell repeatedly in either the x or y direction, a distance d at a time, you could generate the entire array. In the sketch we have used dotted lines to indicate several possible unit cells. All the cells have the same area, and by moving them up or down or across we could locate all the sites in the array. Ordinarily we select the unit cell on the left, because it includes four of the atoms and has its edges along the natural axes, x and y, for the array, but that is not necessary. In fact, the middle cell has the advantage that it clearly tells us that the number of atoms in the cell is equal to 1. The number of atoms in whatever cell we choose must also equal 1, but that is not so apparent in the cell on the left. However, once you realize that only 1⁄4 of each atom on the corners of the cell actually belongs to that cell, because it is shared by three other cells, the number of atoms in the whole cell becomes equal to 1⁄4 × 4, or 1, which is what we got by drawing the cell in a different position.


114

Experiment 16

The Structure of Crystals—An Experiment Using Models

When we extend the array to three dimensions, the same ideas regarding unit cells apply. The unit cell is the smallest portion of the array that could be used to generate the array. With cubic cells the unit cell is usually chosen to have edges parallel to the x, y, and z axes we could put on the array.

Experimental Procedure

IN THIS EXPERIMENT YOUR INSTRUCTOR MAY ALLOW YOU TO WORK WITHOUT SAFETY GLASSES

In this experiment we will mix the discussion with the experimental procedure, since one supports and illustrates the other. We will start with the simplest possible cubic crystal, deal with its properties, and then go on to the next more complex example, and the next, and so on. Each kind of crystal will be related in some ways to the earlier ones but will have its own properties as well. So get out your model set and let’s begin. Work in pairs, and complete each section before going on to the next, unless directed otherwise by your instructor.

The Simple Cubic Crystal You can probably guess the form of the array in the crystal structure we call simple cubic (SC). It is shown on the Calculations page at the end of this section. The unit cell is a cube with an edge length equal to the distance from the center of one atom to the center of the next. The cell edge is usually given the symbol d. The volume of the unit cell is d3, and is very small, since d is of the order of 0.5 nm. Using X-rays we can measure d to four significant figures quite easily. The number of atoms in the unit cell in a simple cubic crystal is equal to 1. Can you see why? Only 1/8 of each corner atom actually belongs to the cell, since it is shared equally by eight cells. Using your model set, assemble three attached unit cells having the simple cubic structure shown in the sketch. Use the short bonds (no. 6) and the gray atoms. Each bond goes into a square face on the atom. An actual crystal with this structure would have many such cells, in three dimensions. If you extended the model you have made further, you would find that each atom would be connected to six others. We say that the coordination number of the atoms in this structure is 6. Only the closest atoms are considered to be bonded to each other. In the other models we will be making, only those atoms that are bonded, by covalent or ionic bonds, will be connected to one another, so the number of bonds to an atom in a large model will be equal to the coordination number. Although the model has an open structure to help us see relationships better, in an actual crystal we consider that the atoms that are closest are touching. It is on this assumption that we determine atomic radii. In this SC crystal, if we know d we can find the atomic radius r of the atoms, since, if the atoms are touching, d must equal 2r. Another property we can calculate knowing d is the density, given the nature of the atoms in the crystal. From d we can easily find the volume V of the unit cell. Since there is one atom per cell, a mole will contain Avogadro’s number of cells. Given the molar mass of the element, we can find the mass m of a cell. The density is simply m/V. With more complex crystals we can make these same calculations, but must take account of the fact that the number of atoms or ions per unit cell may not be equal to one. Essentially no elements crystallize in an SC structure. The reason is that SC packing is inefficient, in that the atoms are farther apart than they need be. There is, as you can see, a big hole in the middle of each unit cell that is begging for an atom to go there, and atoms indeed do. Before we go into that, let’s calculate the fraction of the volume of the unit cell that is actually occupied by atoms. This is easy to do. Make the calculation of that fraction on the Calculations page. That fraction is indeed pretty small; only about 52 percent of the cell volume is occupied by atoms. Most of the empty space is in that hole. You can calculate by simple geometry that an atom having a radius equal to 73 percent of that of the atom on a corner would fit into the hole. Or, putting it another way, an atom bigger than that would have to push the corner atoms back to fit in. If the particles on the corners were anions, and the atom in the hole were a cation, the cation in the hole would push the anions apart, decreasing the repulsion between them and maximizing the attraction between anions and cations. We will have more to say about this when we deal with binary salts, but for now you need to note that if r+/r− > 0.732, the cation will not fit in the cubic hole formed by anions on the corners of an SC cell.


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115

Body-Centered Cubic Crystals In a body-centered cubic (BCC) crystal, the unit cell still contains the corner atoms present in the SC structure, but in the center of the cell there is another atom of the same kind. The unit cell is shown on the Calculations page. Using your model set, assemble a BCC crystal. Use the blue balls. Put the short bonds in the eight holes in the triangular faces on one blue ball. Attach a blue ball to each bond, again using the holes in a triangular face. Those eight blue balls define the unit cell in this structure. Add as many atoms to the structure as you have available, so that you can see how the atoms are arranged. In a BCC crystal each atom is bonded to eight others, so the coordination number is 8. Verify this with your model. (There are no bonds along the edges of the unit cell, since the corner atoms are not as close to one another as they are to the central atom.) The BCC lattice is much more stable than the SC one, in part at least because of the higher coordination number. Many metals crystallize in a BCC lattice—including sodium, chromium, tungsten, and iron—when these metals are at room temperature. There are several properties of the BCC structure that you should note. The number of atoms per unit cell is two, one from the corner atoms and one from the atom in the middle of the cell, where it is unshared. As with SC cells, there is a relation between the unit cell size and the atom radius. Given that in sodium metal the cell edge is 0.429 nm, calculate the radius of a sodium atom on the Calculations page. When you are finished, calculate the density of sodium metal. The fraction of the volume that is occupied by atoms in a BCC crystal is quite a bit larger than with SC crystals. Using the same kind of procedure that we did with the SC structure, we can show that in BCC crystals about 68 percent of the cell volume is occupied.

Close-Packed Structures Although many elements form BCC crystals, still more prefer structures in which the atoms are close packed. In such structures there are layers of atoms in which each atom is in contact with six others, as in the sketch below:

This is the way that billiard balls lie in a rack, or the honeycomb cells are arranged in a bees’ nest. It is the most efficient way one can pack spheres, with about 74 percent of the volume in a close-packed structure filled with atoms. It turns out that there is more than one close-packed crystal structure. The layers all have the same structure, but they can be stacked on one another in two different ways. This is certainly not apparent at first sight. The first close-packed crystal we will examine is cubic, amazingly enough, considering all those triangles in the layers of atoms.

Face-Centered Cubic Crystals In the face-centered cubic (FCC) unit cell there are atoms on the corners and an atom at the center of each face. There is no atom in the center of the cell (see the sketch on the Calculations page, where we show the bonding on only the three exposed faces). Constructing an FCC unit cell, using bonds between closest atoms, is not as simple as you might think, so let’s do it the easy way. This time select the large gray balls, the ones with the most holes. Make the bottom face of the unit cell by attaching four gray balls to a center ball, using short bonds, and holes that are in rectangular faces. You should get the structure shown on the following page:


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x x

x

x x

x

x

x x

To make the middle layer of atoms, connect four of the balls in a square, again using short bonds and the holes in rectangular faces, as in the right sketch. The top layer is made the same way as the bottom one. Then connect the layers to one another, again using holes in the rectangular faces. You should then have an FCC cubic unit cell. Because the atoms in the middle of adjacent faces are as close to one another as they are to the corner atoms, there should be bonds between them. There are quite a few bonds in the final cell, 32 in all. Put them all in. As with BCC there are no bonds along the edges of the cell. The FCC, or cubic close-packed, structure is a common one. Among the metals with this type of crystal are copper, silver, nickel, and calcium. In close-packed structures the coordination number is the largest that is possible. It can be found by looking at a face-centered atom, say on the top face of the cell. That atom has eight bonds to it and would also have bonds to atoms in the cell immediately above, four of them. So its coordination number, like that of every atom in the cell, is 12. Having seen how to deal with other structures, you should now be able to find the number of atoms in the FCC unit cell and the radius of an atom as related to d. From these relationships and the measured density, you can calculate Avogadro’s number. On the Calculations page, find these quantities, using copper metal, which has a unit cell edge d equal to 0.361 nm and a density equal to 8.92 g/cm3. In the center of the unit cell there is a hole. It is smaller than a cubic hole but of significant size nonetheless. It is called an octahedral hole, because the six atoms around it define an octahedron. In an ionic crystal, the anions often occupy the sites of the atoms in your cell, and there is a cation in the center of the octahedral hole. On the Calculations page, find the maximum radius, r+, of the cation that would just fit in an octahedral hole surrounded by anions of radius r−. You should find that the r+/r− ratio turns out to be 0.414. There is another kind of hole in the FCC lattice that is important. See if you can find it; there are eight of them in the unit cell. If you look at an atom on the corner of the cell, you can see that it lies in a tetrahedron. In the center of the tetrahedron there is a tetrahedral hole, which is small compared to an octahedral or cubic hole. However, in some crystals small cations are found in some or all of the tetrahedral holes, again in a closepacked anion lattice. The cation-anion radius ratio at which cations would just fit into a tetrahedral hole is not so easy to find; r+/r− turns out to be 0.225. The close-packed layers of atoms in the FCC structure are not parallel to the unit cell faces, but rather are perpendicular to the cell diagonal. If you look down the cell diagonal, you see six atoms in a close-packed triangle in the layer immediately behind the corner atom, and another layer of close-packed atoms below that, followed by another corner atom. The layers are indeed closely packed, and, as one goes down the diagonal of this and succeeding cells, the layers repeat their positions in the order ABCABC…, meaning that atoms in every fourth layer lie below one another. Clearly there is another way we could stack the layers. The first and second will always be in the same relative positions, but the third layer could be below the first one if it were shifted properly. So we can have a closepacked structure in which the order of the layers is ABABAB…. The crystal obtained from this arrangement of layers is not cubic, but hexagonal. It too is a common structure for metals. Cadmium, zinc, and manganese have this structure. As you might expect, the stability of this structure is very similar to that of FCC crystals. We find that simply changing the temperature often converts a metal from one form to another. Calcium, for example, is FCC at room temperature, but if heated to 450°C it converts to close-packed hexagonal.

Crystal Structures of Some Common Binary Compounds We have now dealt with all of the possible cubic crystal structures for metals. It turns out that the structures of binary ionic compounds are often related to these metal structures in a very simple way. In many ionic crystals the anions, which are large compared with cations, are essentially in contact with each other, in either an


Experiment 16


117

SC or FCC structure. The cations go into the cubic, or octahedral, or tetrahedral holes, depending on the cation-anion radius ratios that we calculated. The idea is that the cation will tend to go into a hole in which it will not quite fit. This increases the unit cell size from the value it would have if the anions were touching, which reduces the repulsion energy due to anion-anion interaction and increases to the maximum the cationanion attraction energy, producing the most stable possible crystal structure. According to the so-called radiusratio rule, large cations go into cubic holes, smaller ones into octahedral holes, and the smallest ones into tetrahedral holes. The deciding factor for which hole is favored is given by the radius ratio: If r+/r− > 0.732 If 0.732 > r+/r− > 0.414 If 0.414 > r+/r− > 0.225

cations go into cubic holes cations go into octahedral holes cations go into tetrahedral holes

The NaCl Crystal To apply the radius-ratio rule to NaCl, we simply need to find r+/r−, using the data in Table 16.1. Since rNa+ = 0.095 nm, and rCl− = 0.181 nm, the radius ratio is 0.095/0.181, or 0.525. That value is less than 0.732 and greater than 0.414, so the sodium ions should go into octahedral holes. We saw the octahedral hole in the center of the FCC unit cell, so Na+ ions go there, and the Cl− ions have the FCC structure. Actually, there are 12 other octahedral holes associated with the cell, one on each edge, which would be apparent if we had been able to make more cells. An Na+ ion goes into each of these holes, giving the classic NaCl structure shown on the Calculations page. (Again, only the ions and bonds on the exposed faces are shown.) Make a model of the unit cell for NaCl, using the large gray balls for the Cl− ions and the blue ones for + Na ions. Clearly the Na+ ion at the center of the cell is in an octahedral hole, but so are all of the other sodium ions, because if you extend the lattice, every Na+ ion will be surrounded by six Cl− ions. The coordination number of Na+, and of Cl−, ions is 6. The crystal is FCC in Cl− and also in Na+, because you could put Na+ ions on the corners of the unit cell and maintain the same structure. The unit cell extends from the center of one Cl− ion to the center of the next Cl− along the cell edge; or, from the center of one Na+ ion to the center of the next Na+ ion. On the Calculations page find the number of Na+ and of Cl− ions in the unit cell.

The CsCl Crystal Cesium chloride has the same type of formula as NaCl, 1:1. The Cs+ ion, however, is larger than Na+, and has a radius equal to 0.169 nm. This makes r+/r− equal to 0.933. Because this value is greater than 0.732, we would expect that in the CsCl crystal the Cs+ ions will fill cubic holes, and this is what is observed. The structure of CsCl will look like that of the BCC unit cell you made earlier, except that the ion in the center will be Cs+ and those on the corners Cl−. If you put gray balls in the center of each BCC unit cell you made from the blue balls, you would have the CsCl structure. This structure is not BCC, because the corner and center atoms are not the same. Rather it consists of two interpenetrating simple cubic lattices, one made from Cl− ions and the other from Cs+ ions.

The Zinc Sulfide Crystal Zinc sulfide is another 1:1 compound, but its crystal structure is not that of NaCl or of CsCl. In ZnS, the Zn2+ ions have a radius of 0.074 nm and the S2− ions a radius of 0.184 nm, making r+/r− equal to 0.402. By the radius-ratio rule, ZnS should have close-packed S2− ions, with the Zn2+ ions in tetrahedral holes. In the ZnS unit cell, we find that this is indeed the case; the S2− ions are FCC, and alternate tetrahedral holes in the unit cell are occupied by Zn2+ ions, which themselves form a tetrahedron. To make a model for ZnS, first assemble an SC unit cell, using blue balls (b) and the long bonds. Use gray balls (g) for the zinc ions, and assemble the unit shown on the following page, using the short bonds and the holes in triangular faces:


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g

g

b

Attach this unit to two corner atoms on the diagonal of the bottom face. Make another unit and attach it to the two corner atoms in the upper face that lie on the face diagonal that is not parallel to the bottom face diagonal. The gray balls should then form a tetrahedron. Then attach the four other FCC blue balls to the gray ones. In this structure the coordination number is four (the long bonds do not count, they just keep the unit cell from falling apart). If the gray balls in the unit cell are replaced with blue ones, so that all atoms are of the same element, we obtain the diamond crystal structure. These are the three common cubic structures of 1:1 compounds. The radius-ratio rule allows us to predict which structure a given compound will have. It does not always work, but it is correct most of the time. On the Calculations page, use the rule to predict the cubic structures that crystals of the following substances will have: KI, CuBr, and TlBr.

The Calcium Fluoride Crystal Calcium fluoride is a 1:2 compound, so it cannot have the structure of any crystal we have discussed so far. The radii of Ca2+ and F − are 0.099 and 0.136 nm, respectively, so r+/r− is 0.727. This makes CaF2 on the boundary between compounds with cations in cubic holes or octahedral holes. It turns out that in CaF2, the F − ions have a simple cubic structure, with half of the cubic holes filled by Ca2+ ions. This produces a crystal in which the Ca2+ ions lie in an FCC lattice, with 8 F − ions in a cube inside the unit cell. Use your model set to make a unit cell for CaF2. Use gray balls (g) for the Ca2+ ions and red ones (r) for F−. The cations and anions are linked by short bonds that go into holes in nonadjacent triangular faces. The Ca2+ ions on each face diagonal are linked to F − ions as shown below: r

g

r

g

g

To complete the bottom face, attach two red and two gray balls to the initial line, so that the four red balls form a square and the gray ones an FCC face. The top of the unit cell is made the same way. Attach the two assemblies through four gray balls, which lie at the centers of the other faces. When you are done, the red balls should form a cube inside an FCC unit cell made from gray balls. Extend the model into an adjacent unit cell to show that every other cube of F− ions has a Ca2+ ion at its center. When you have completed this part of the experiment, your instructor may assign you a model with which to work. Report your results as directed. Then disassemble the models you made and pack the components in the box.


Name ____________________________________

Section _________________________________

Experiment 16

Data and Calculations: The Structure of Crystals Table 16.1 Atom Na Cu Cl Cs Zn S K I Br Tl

Molar Mass, g/mole 22.99 63.55 35.45 132.9 65.38 32.06 39.10 126.9 79.90 204.4

Atomic Radius, nm

Ionic Radius, nm

— — 0.099 0.262 0.133 0.104 0.231 0.133 0.114 0.171

0.095 0.096 0.181 0.169 0.074 0.184 0.133 0.216 0.195 0.147

Some Cubic Unit Cells

Simple Cubic Crystal Fraction of volume of unit cell occupied by atoms Volume of unit cell in terms of d

____________

Number of atoms per unit cell

____________

Radius r of atom in terms of d

____________

4π ⎞ ⎛ Volume of atom in terms of d ⎝ V = 3 r 3 ⎠

____________

Volume of atom/volume of cell

____________ = ____________ (continued on following page)


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Body-Centered Cubic Crystal a.

Radius of a sodium atom In BCC atoms touch along the cube diagonal (see your model) Length of cube diagonal if d = 0.429 nm

____________ nm

4 × rNa = length of cube diagonal b.

rNa = ____________ nm

Density of sodium metal Length of unit cell, d, in cm (1 cm = 107 nm)

____________ cm

Volume of unit cell, V

____________ cm3


____________

Number of unit cells per mole Na

____________

Mass of a unit cell, m

____________ g

Density of sodium metal, m/V (Observed value = 0.97 g/cm3)

____________ g/cm3

Face-Centered Cubic Crystal a.

Number of atoms per unit cell Number of atoms on corners ________ Shared by ________ cells ________ × ________ = ________ Number of atoms on faces ________ Shared by ________ cells

________ × ________ = ________ Total atoms per cell ____________

b.

Radius of a Cu atom In FCC atoms touch along face diagonal (see your model and ASA) Length of face diagonal in Cu, where d = 0.361 nm Number of Cu atom radii on face diagonal _____

____________ nm rCu = ____________ nm (continued on following page)


Experiment 16

c.

d.


121

Avogadro’s Number, N Length of cell edge, d, in cm, in Cu metal

____________ cm

Volume of a unit cell in Cu metal

____________ cm3

Volume of a mole of Cu metal (V = mass/density)

____________ cm3

Number of unit cells per mole Cu

____________


____________

Number of atoms per mole, N

____________

Size of an octahedral hole Below is a sketch of the front face of a face-centered cubic cell with the NaCl structure. The ions are drawn so that they are just touching their nearest neighbors. The cations are in octahedral holes.

Show the length of d on the sketch. (See your model.) What is the relationship between r− and d? r− = ____________ × d What is the equation relating r+, r−, d? d = ____________ What is the relationship between r+ and d? r+ = ____________ × d What is the value of the radius ratio r+/r−? Express the ratio to three significant figures. r+/r− = ____________ (continued on following page)


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Experiment 16


NaCl Crystal Number of Cl− ions in the unit cell (FCC)

____________

Number of Na+ ions on edges of cell ____________ Shared by ____________ cells Number of Na+ ions in center of cell ____________ Shared by ____________ cell(s) Total no. Na+ ions in unit cell

____________

Radius-ratio rule (data in Table 16.1) KI

r+/r− ____________ Predicted structure ____________ Obs. NaCl

CuBr

r+/r− ____________ Predicted structure ____________ Obs. ZnS

TlBr

r+/r− ____________ Predicted structure ____________ Obs. CsCl

Report on unknown


Name ____________________________________

Section _________________________________

Experiment 16

Advance Study Assignment: The Structure of Crystals 1. Many substances crystallize in a cubic structure. The unit cell for such crystals is a cube having an edge with a length equal to d.

a. What is the length, in terms of d, of the face diagonal, which runs diagonally across one face of the cube? ____________ b. What is the length, again in terms of d, of the cube diagonal, which runs from one corner, through the center of the cube, to the other corner? Hint: Make a right triangle having a face diagonal and an edge of the cube as its sides, with the hypotenuse equal to the cube diagonal. ____________ 2. In an FCC structure, the atoms are found on the corners of the cubic unit cell and in the centers of each face. The unit cell has an edge whose length is the distance from the center of one corner atom to the center of another corner atom on the same edge. The atoms on the diagonal of any face are touching. One of the faces of the unit cell is shown below.

a. Show the distance d on the sketch. Draw the boundaries of the unit cell. b. What is the relationship between the length of the face diagonal and the radius of the atoms in the cell? Face diagonal = ____________ (continued on following page)


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Experiment 16


c. How is the radius of the atoms related to d?

r = ____________ d. Silver metal crystals have an FCC structure. The unit cell edge in silver is 0.4086 nm long. What is the radius of a silver atom? ____________nm


Experiment 17 Classification of Chemical Substances

epending on the kind of bonding present in a chemical substance, the substance may be called ionic, molecular, or metallic. In a solid ionic compound there are ions; the large electrostatic forces between the positively and negatively charged ions are responsible for the bonding which holds these particles together. In a molecular substance the bonding is caused by the sharing of electrons by atoms. When the stable aggregates resulting from covalent bonding contain relatively small numbers of atoms, they are called molecules. If the aggregates are very large and include essentially all the atoms in a macroscopic particle, the substance is called macromolecular. Metals are characterized by a kind of bonding in which the electrons are much freer to move than in other kinds of substances. The metallic bond is stable but is probably less localized than other bonds. The terms ionic, molecular, macromolecular, and metallic are somewhat arbitrary, and some substances have properties that would place them in a borderline category, somewhere between one group and another. It is useful, however, to consider some of the general characteristics of typical ionic, molecular, macromolecular, and metallic substances, since many very common substances can be readily assigned to one of these categories or another.

D

Ionic Substances Ionic substances are all solids at room temperature. They are typically crystalline but may exist as fine powders as well as clearly defined crystals. While many ionic substances are stable up to their melting points, some decompose on heating. It is very common for an ionic crystal to release loosely bound water of hydration at temperatures below 200°C. Anhydrous (dehydrated) ionic compounds have high melting points, usually above 300°C but below 1000°C. They are not readily volatilized and boil only at very high temperatures (Table 17.1). When molten, ionic compounds conduct an electric current. In the solid state they do not conduct electricity. The conductivity in the molten liquid is attributed to the freedom of motion of the ions, which arises when the crystal lattice is no longer present. Ionic substances are frequently but not always appreciably soluble in water. The solutions produced conduct the electric current rather well. The conductivity of a solution of a slightly soluble ionic substance is often several times that of the solvent water. Ionic substances are usually not nearly so soluble in other liquids as they are in water. For a liquid to be a good solvent for ionic compounds it must be highly polar, containing molecules with well-defined positive and negative regions with which the ions can interact.

Molecular Substances All gases and essentially all liquids at room temperature are molecular in nature. If the molar mass of a molecular substance is over about 100 grams, it may be a solid at room temperature. The melting points of molecular substances are usually below 300°C; these substances are relatively volatile, but a good many will decompose before they boil. Most molecular substances do not conduct the electric current either when solid or when molten. Organic compounds, which contain primarily carbon and hydrogen, often in combination with other nonmetals, are essentially molecular in nature. Since there are a great many organic substances, it is true that most substances are molecular. If an organic compound decomposes on heating, the residue is frequently a


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Experiment 17

Classification of Chemical Substances

Table 17.1 Physical Properties of Some Representative Chemical Substances Solubility Substance NaCl MgO CoCl2 CoCl2 ⋅ 6H2O C10H8 C6H5COOH FeCl3 SnI4 SiO2 Fe

M.P., °C

B.P., °C

Water

Hexane

1413

Sol

Insol

— 1049 Dec

Sl sol Sol Sol

Insol Insol Insol

70 122 282

255 249 315

Insol Sl sol Sol

Sol Sl sol Sl sol

144 1600 1535

341 2590 3000

Dec Insol Insol

Sol Insol Insol

801 2800 Sublimes 86

Electrical Conductance High in melt and in soln Low in sat’d soln High in soln High in soln Zero in melt Low in sat’d soln High in soln and in melt ∼Zero in melt Zero in solid High in solid

Classification Ionic Ionic Ionic Ionic hydrate, −H2O at 110°C Molecular Molecular-ionic Molecular-ionic Molecular Macromolecular Metallic

Key: Sol = at least 0.1 mole/L; Sl sol = appreciable solubility but 20,000Ω.

______________

______________

Unknown no.

VI

V

IV

III

II

I

Substance No.

Table 17.2

Observations and Conclusions: Classification of Chemical Substances

Experiment 17

Name ____________________________________

Classification and Reason


Name ____________________________________

Section ________________________________

Experiment 17

Advance Study Assignment: Classification of Chemical Substances 1.

List the properties of a substance that would definitely establish that the material is molecular.

2.

If we classify substances as ionic, molecular, macromolecular, or metallic, in which if any categories are all the members a. soluble in water?

b. electrical conductors in the melt?

c. insoluble in all common solvents?

d. solids at room temperature?

3.

A given substance is a white, granular solid at 25°C that does not conduct electricity. It melts at 750°C without decomposing, and the melt conducts an electric current. It dissolves readily in water, to produce an electrically conductive solution. What would be the classification of the substance, based on this information?

4.

A lustrous grey-white solid melts at 1650°C. It is electrically conductive as both a solid and a liquid, but not soluble in either water or any organic solvent. Classify the substance as best you can from these properties.



Experiment 18 Some Nonmetals and Their Compounds— Preparations and Properties

ome of the most commonly encountered chemical substances are nonmetallic elements or their simple compounds. O2 and N2 in the air, CO2 produced by combustion of oil, coal, or wood, and H2O in rivers, lakes, and air are typical of such substances. Substances containing nonmetallic atoms, whether elementary or compound, are all molecular, reflecting the covalent bonding that holds their atoms together. They are often gases, due to weak intermolecular forces. However, with high molecular masses, hydrogen bonding, or macromolecular structures one finds liquids, like H2O and Br2, and solids, like I2 and graphite. Several of the common nonmetallic elements and some of their gaseous compounds can be prepared by simple reactions. In this experiment you will prepare some typical examples of such substances and examine a few of their characteristic properties.

S

Experimental Procedure In several of the experiments you perform, you will prepare gases. In general we will not describe in detail should be observed, so perform each preparation carefully and report what you actually observe, not what you think we expect you to observe. There are several tests we will make on the gases you prepare, and the way each of these should be carried out is summarized below. To determine the odor of a gas, first pass your hand across the end of the tube, bringing the gas toward your nose. If you don’t detect an odor, sniff near the end of the tube, first at some distance and then gradually closer. Don’t just put the tube at the end of your nose immediately and take a deep breath. Some of the gases you will make have no odor, and some will have very impressive ones. Some of the gases are very toxic and, even though we will be making only small amounts, caution in testing for odor is important. Test for Odor.

A few gases will support combustion, but most will not. To make the test, ignite a wood splint with a Bunsen flame, blow out the flame, and put the glowing, but not burning, splint into the gas in the test tube. If the gas supports combustion, the splint will glow more brightly, or may make a small popping noise. If the gas does not support combustion, the splint will go out almost instantly. You can use the same splint for all the combustion tests.

Test for Support of Combustion.

Many gases are acids. This means that if the gas is dissolved in water it will produce some H ions. A few gases are bases; in water solution such gases produce OH− ions. Other gases do not interact with water and are neutral. It is easy to establish the acidic or basic nature of a gas by using a chemical indicator. One of the most common acid-base indicators is litmus, which is red in acidic solution and blue in basic solution. To test whether a gas is an acid, moisten a piece of blue litmus paper with water from your wash bottle, and put the paper down in the test tube in which the gas is present. If the gas is an acid, the paper will turn red. Similarly, to test if the gas is a base, moisten a piece of red litmus paper, and hold it down in the tube. A color change to blue will occur if the gas forms a basic solution. Since you may have used an acid or a base in making the gas, do not touch the walls of the test tube with the paper. The color change will occur fairly quickly and smoothly over the surface of the paper if the gas is acidic or basic. It is not necessary to use a new piece of litmus for each test. Start with a piece of blue and a piece of red litmus. If you need to Test for Acid-Base Properties. +


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Experiment 18

Some Nonmetals and Their Compounds—Preparations and Properties

regenerate the blue paper, hold it over an open bottle of 6 M NH3, whose vapor is basic. If you need to make red litmus, hold the moist paper over the acidic vapor of an open bottle of 6 M acetic acid.

A. Preparation and Properties of Nonmetallic Elements: O2, N2, Br2, I2 1. Oxygen, O2. Oxygen can be easily prepared in the laboratory by the decomposition of H2O2, hydrogen peroxide, in aqueous solution. Hydrogen peroxide is not very stable and will break down to water and oxygen gas on addition of a suitable catalyst, particularly MnO2:

2 H 2O2 (aq ) ⎯MnO ⎯⎯2 → O2 (g) + 2 H 2O

(1)

Add 1 mL 3% H2O2 solution in water to a small test tube. Pick up a small amount (∼0.1 g) of MnO2 on the tip of your spatula and add it to the liquid in the tube. Hold your finger over the end of the tube to help confine the O2. Test the evolved gas for odor. Test the gas for any acid-base properties, using moist blue and red litmus paper. Test the gas for support of combustion. Record your observations. 2. Nitrogen, N2.

Sodium sulfamate in water solution will react with nitrite ion to produce nitrogen gas: NO2 − (aq) + NH 2SO3− (aq) → N 2 (g) + SO 4 2− (aq) + H 2O

(2)

Add about 1 mL 1 M KNO2, potassium nitrite, to a small test tube. Then add 10 to 12 drops 0.5 M NaNH2SO3, sodium sulfamate, and place the test tube in a hot-water bath made from a 250-mL beaker half full of water. Bubbles of nitrogen should form within a few moments. Confine the gas for a few seconds with a stopper. Cautiously test the evolved gas for odor. Carry out the tests for acid-base properties and for support of combustion. If you need to generate more N2 to complete the tests, add sodium sulfamate solution as necessary, 5 to 10 drops at a time. Report your observations. 3. Iodine, I2. The halogen elements are most readily prepared from their sodium or potassium salts. The reaction involves an oxidizing agent, which can remove electrons from the halide ions, freeing the halogen. The reaction that occurs when a solution of potassium iodide, KI, is treated with 6 M HCl and a little MnO2 is

2 I − (aq) + 4 H + (aq) + MnO2 (s) → I 2 (aq) + Mn 2+ (aq) + 2 H 2O

(3)

At 25°C I2 is a solid, with relatively low solubility in water, and appreciable volatility. I2 can be extracted from aqueous solution into organic solvents, particularly heptane, C7H16 (HEP). The solid, vapor, and solutions in water and HEP all have characteristic colors. Put two drops 1 M KI into a regular (18 × 150 mm) test tube. Add six drops 6 M HCl and a tiny amount of manganese dioxide, MnO2. Swirl the mixture and note any changes that occur. Put the test tube into a hotwater bath. After a minute or two a noticeable amount of I2 vapor should be visible above the liquid. Remove the test tube from the water bath and add 10 mL distilled water. Stopper the tube and shake. Note the color of I2 in the solution. Sniff the vapor above the solution. Decant the liquid into another regular test tube and add 3 mL heptane, C7H16. Stopper and shake the tube. Observe the color of the HEP layer and the relative solubility of I2 in water and HEP. Record your observations. 4. Bromine, Br2. Bromine can be made by the same reaction as is used to make iodine, substituting bromide ion for iodide. Bromide ion is less easily oxidized than iodide ion. Bromine at 25°C is a liquid. Br2 can be extracted from water solution into heptane. Its liquid, vapor, and solutions in water and HEP are colored. To two or three drops 1 M NaBr in a regular test tube, add six drops 6 M HCl and a small amount of MnO2, about the size of a small pea. Swirl to mix the reagents, and observe any changes. Heat the tube in the water bath for a minute or two. Try to detect Br2 vapor above the liquid by observing its color. Add 10 mL water to the tube, stopper, and shake. Note the color of Br2 in the solution. Sniff the vapor, cautiously. Decant the liquid into a regular test tube. Add 3 mL HEP, stopper, and shake. Note the color of Br2 in HEP. Record your observations.


Experiment 18


135

B. Preparation and Properties of Some Nonmetallic Oxides: CO2, SO2, NO, and NO2 1. Carbon Dioxide, CO2. Carbon dioxide, like several nonmetallic oxides, is easily made by treating an oxyanion with an acid. With carbon dioxide the oxyanion is CO3 2− , carbonate ion, which is present in solutions of carbonate salts, such as Na2CO3. The reaction is

CO32− (aq) + 2 H + (aq) → (aq) → H 2CO3 (aq) → CO2 (g) + H 2O

(4)

Carbon dioxide is not very soluble in water, and on acidification, carbonate solutions will tend to effervesce as CO2 is liberated. Nonmetallic oxides in solution are often acidic but never basic. To 1 mL M Na2CO3 in a small test tube add six drops 3 M H2SO4. Test the gas for odor, acidic properties, and ability to support combustion. 2. Sulfur Dioxide, SO2. Sulfur dioxide is readily prepared by acidification of a solution of sodium sulfite, containing sulfite ion, SO32 − . As you can see, the reaction that occurs is very similar to that with carbonate ion.

SO32− (aq) + 2 H + (aq) → H 2SO3 (aq) → SO2 (g) + H 2O

(5)

Sulfur dioxide is considerably more soluble in water than is carbon dioxide. Some effervescence may be observed on acidification of concentrated sulfite solutions; effervescence will increase if the solution is heated in a water bath. To 1 mL M Na2SO3 in a small test tube add six drops 3 M H2SO4. Cautiously test the evolved gas for odor. Test its acidic properties and its ability to support combustion. Put the test tube into the water bath for a few seconds to see if effervescence occurs if the solution is hot. If a solution containing nitrite ion, NO 2 − , is treated with acid, two oxides are produced, NO2 and NO. In solution these gases are combined in the form of N2O3, which is colored. When these gases come out of solution, the mixture contains NO and NO2; the latter is colored. NO is colorless and reacts readily with oxygen in the air to form NO2. The preparation reaction is 3. Nitrogen Dioxide, NO2, and Nitric Oxide, NO.

2 NO2 − (aq) + 2 H + (aq) → N 2O3 (aq) + H 2O → NO(g) + NO2 (g) + H 2O

(6)

To 1 mL of 1 M KNO2 in a small test tube, add six drops 3 M H2SO4. Swirl the mixture for a few seconds and note the color of the solution. Warm the tube in the water bath for a few seconds to increase the rate of gas evolution. Note the color of the gas that is given off. Cautiously test the odor of the gas. Test its acidic properties and its ability to support combustion. Record your observations.

C. Preparation and Properties of Some Nonmetallic Hydrides: NH3, H2S A 6 M solution of NH3 in water is a common laboratory reagent. You may have been using it in this experiment to make your litmus paper turn blue. Ammonia gas can be made by simply heating 6 M NH3. It can also be prepared by addition of a strongly basic solution to a solution of an ammonium salt, such as NH4Cl. The latter solution contains NH 4 + ion. On treatment with OH− ion, as in a solution of NaOH, the following reaction occurs: 1. Ammonia, NH3.

NH 4 + (aq) + OH − (aq) → NH3 (aq) + H 2O → NH3 (g) + H 2O

(7)

The odor of NH3 is characteristic. NH3 is very soluble in water, so effervescence is not observed, even on heating concentrated solutions. Add 1 mL 1 M NH4Cl to a small test tube. Add 1 mL 6 M NaOH. Swirl the mixture and test the odor of the evolved gas. Put the test tube into the water bath for a few moments to increase the amount of NH3 in the gas phase. Test the gas with moistened blue and red litmus paper. Test the gas for support of combustion.


136

Experiment 18


2. Hydrogen Sulfide, H2S. Hydrogen sulfide can be made by treating some solid sulfides, particularly FeS, with an acid such as HCl or H2SO4. With FeS the reaction is

FeS(s) + 2 H + (aq) → H 2S(g) + Fe 2+ (aq)

(8)

This reaction was used for many years to make H2S in the laboratory in courses in qualitative analysis. In this experiment we will employ the method currently used in such courses for H2S generation. This involves the decomposition of thioacetamide, CH3CSNH2, which occurs in solution on treatment with acid and heat. The reaction is CH3CSNH 2 (aq) + 2 H 2O → H 2S(g) + CH3COO − (aq) + NH 4 + (aq)

(9)

The odor of H2S is notorious. H2S is moderately soluble in water and, since it is produced reasonably slowly in Reaction 9, there will be little if any effervescence. To 1 mL of 1 M thioacetamide in a small test tube add six drops 3 M H2SO4. Put the test tube in a boilingwater bath for about 1 minute. There may be some cloudiness due to formation of free sulfur. Carefully smell the gas in the tube; H2S is toxic. Test the gas for acidic and basic properties and for the ability to support combustion.

D. Identification of Unknown Solution

Optional

In this experiment you prepared nine different species containing nonmetallic elements. In each case the source of the species was in solution. In this part of the experiment we will give you an unknown solution that can be used to make one of the nine species. It will be a solution used in preparing one of the species, but it will not be an acid or a base. Identify by suitable tests the species that can be made from your unknown and the substance that is present in your unknown solution. CAUTION:

The following gases prepared in this experiment are toxic: Br2, SO2, NO2, NH3, and H2S. Do not inhale these gases unnecessarily when testing their odors. One small sniff will be sufficient and will not be harmful. It is good for you to know these odors, in case you encounter them in the future. Most of the chemicals used in this experiment can be discarded down the sink drain. Pour the solutions from Sections 3 and 4 of Part A, containing I2 and Br2 in heptane, into a waste crock, unless directed otherwise by your instructor. DISPOSAL OF REACTION PRODUCTS.


Name _____________________________________ Section _________________________________

Experiment 18

Observations and Conclusions: Some Nonmetals and Their Compounds A.

Element Prepared

Degree of Effervescence

Odor

Acid-Base Tests

Support of Combustion Test

1.

O2

______________ ______________

______________

______________

2.

N2

______________ ______________

______________

______________

3.

I2

Color _________________________________________________ In HEP Odor Vapor In H2O ______________ ______________ ______________ ______________

4.

Br2

B.

Oxide Prepared

______________ ______________


Odor

______________

Acid Test

______________ Support of Combustion Test

Color

1.

CO2

______________ ______________ ______________ ______________ ______________

2.

SO2

______________ ______________ ______________ ______________ ______________

soln ______________ 3. NO2 + NO ______________ ______________ ______________ ______________ gas ______________ C.

Hydride Prepared


Odor

Acid-Base Tests

Support of Combustion Test

1.

NH3

______________ ______________

______________

______________

2.

H2S

______________ ______________

______________

______________

D. Properties of Unknown Solution

Optional

Nonmetal species that can be made from unknown ______________ Identity of unknown solution ______________ Unknown no. ______________



Name _____________________________________ Section _________________________________

Experiment 18

Advance Study Assignment: Some Nonmetals and Their Compounds In this experiment nine species are prepared and studied. For each of the species, list the reagents that are used in its preparation and the reaction that occurs. Reagents Used 1.

O2

2.

N2

3.

I2

4.

Br2

5.

CO2

6.

SO2

7.

NO2 + NO

8.

NH3

9.

H2S

Reaction



Experiment 19 Molar Mass Determination by Depression of the Freezing Point

he most common liquid we encounter in our daily lives is water. In this experiment we will study the equilibria that can exist between pure water and its aqueous solutions, and ice, the solid form of water. (Water is one of very few substances for which we have a separate name for the solid.) If we take some ice cubes from the refrigerator and put them into a glass of water from the tap, we find that the water temperature falls and some ice melts. This occurs because heat will always tend to flow from a higher to a lower temperature. Heat from the water flows into the ice. It takes heat, called the heat of fusion, to melt ice. If there is enough ice present, the water temperature will ultimately fall to 0°C, and stay there. At that point, ice and water are in equilibrium, at the freezing point of water. At the freezing point, Tf°, the vapor pressures of ice and water must be equal, and that condition fixes the temperature. (See Fig. 19.1.) Now let us consider what happens if we add a soluble liquid or solid to the equilibrium mixture of ice and water. Rather surprisingly, we find that the temperature of the ice and the solution falls as equilibrium is reestablished. The reason this happens is that in the solution the vapor pressure of water at 0°C is less than that of the pure liquid. So the vapor pressure of ice at 0°C is higher than that of the solution, and some ice melts. This requires heat, which comes from the solution and from the ice, and the temperature falls. The vapor pressures of the ice and the solution both fall, but that of the ice falls faster, and at some temperature Tf below 0°C equilibrium is established at the new freezing point. The situation is shown in Figure 19.1. The change in the freezing point that is observed is called the freezing point depression, ΔTf , equal to Tf° − Tf . It is observed with solutions of any solvent. The freezing point depression is one of the colligative properties of solutions. Others are the boiling point elevation for non-volatile solutes, the osmotic pressure, and the vapor pressure lowering. The colligative properties of solutions depend on the number of solute particles present in a given amount of solvent and not on the kinds of particles dissolved, be they molecules, atoms, or ions. When working with colligative properties it is convenient to express the solute concentration in terms of its molality m as defined by the equation:

T

molality of A = mA =

moles A dissolved kg solvent in the solution

(1)

For this unit of concentration, the boiling point elevation, Tb − T b°, or ΔTb, and the freezing point depression, T f° − Tf , or ΔTf , in °C at very low concentrations are given by the equations: ΔTb = kbm

and

ΔTf = kf m

(2)

where kb and kf are characteristic of the solvent used. For water, kb = 0.52 and kf = 1.86. For benzene, kb = 2.53 and kf = 5.10. In this experiment we will assume that Equation 2 is valid, even though our solutions are moderately concentrated. One of the classic uses of colligative properties was in connection with finding molar masses of unknown substances. With organic molecules, molar masses by FP depression agreed with those found by other methods. With ionic salts, like NaCl, molar masses were lower than the formula masses. On the basis of such experiments, Arrhenius suggested that ionic substances exist as ions in aqueous solution, consistent with the observation that such solutions conduct an electric current. Arrhenius’ general idea turned out to be correct, and is now a basic part of modern chemical theory.


142

Experiment 19

Molar Mass Determination by Depression of the Freezing Point

Pure H2O

Vapor pressure mmHg

4.579

Solution ΔTƒ

Freezing point of water

T°ƒ

Freezing point of solution

Tƒ

Ice

0°C 273K

Temperature

Figure 19.1

Discussion of the Method In this experiment we will study the freezing point behavior of some aqueous solutions. First you will measure the freezing point of pure water, using a slurry of ice and water. Then you’ll add a known mass of an unknown to the slurry, and find the freezing point of the solution, thus determining the freezing point depression, ΔTf . That allows you to find the molality of the solution. Finally, you will separate the solution from the ice in the mixture, and weigh the solution, which will contain all of the solute. This information furnishes you with the composition of the solution, and lets you calculate the mass of solute in 1 kg of water. The molar mass follows from Equations 1 and 2.

Experimental Procedure You may work in pairs on this experiment. Both members of the pair must submit a report. From the stock room obtain a digital thermometer, an insulated cup, and two unknowns, one liquid and one solid. The actual molar mass of the solid will be furnished to you.

A. Finding the Freezing Point of Water You will first need to determine the freezing point of pure water. Prepare a water-ice mixture in your insulated cup, using more ice than water. Stir well, and record the lowest temperature you observe for that mixture. (Since your thermometer may not be properly calibrated, that temperature may not be 0.0°C.)

B. Finding the Freezing Point of a Solution of Liquid Unknown To the water-ice mixture you need to add a known amount of one of your unknowns. Let’s work first with the liquid unknown. To estimate the mass of your liquid to add to the mixture, assume that you will have about 100 g of water in the final solution, that the molality of the solute will be about 2 m, and that the molar mass


Experiment 19

Molar Mass Determination by Depression of the Freezing Point

143

of the solute is 50 g. Tare the insulated cup on the top loading-balance, and add the estimated mass by pouring it down a stirring rod on to the center of the ice-water mix. Record the mass. Then record the lowest temperature you observe after stirring the ice + solution mixture. You should obtain a freezing point depression of at least 4°C. If you need to, add more of the liquid unknown, weighing the amount added. Stir well and measure the lowest temperature of the mixture that you observe. Then promptly separate the solution from the ice by pouring it through a wire screen into a tared beaker on the top-loading balance. (Place the screen over the cup, and pour out the solution, keeping a corner of the screen down.) Record the mass of the solution. Make a second molar mass measurement for your liquid unknown. This is easily done if you pour the decanted solution back into the cup once you have found its mass. Add some water and ice to the ice + solution mixture, thereby lowering the molality of the solute. Measure the new freezing point for that mixture. Pour off and weigh the solution. When you have finished this part of the experiment, dispose of the mixture of ice and solution as directed by your instructor. Determine molar masses as described in the Discussion.

C. Finding the Freezing Point of a Solution of Solid Unknown With solid unknowns, you can use the same general procedure as with liquids. First, make a solution of the solid in water, using the same total mass of solid as you did with the liquid. You can do this by taring a small dry beaker on the top-loading balance, and adding the solid, noting its mass. Add a minimum amount of water to dissolve the solid (use at least 20 g of water). Stir until dissolution is complete. Add the entire solution of the solid to a new ice + water mix, noting the lowest temperature you obtain. Decant and weigh the solution as you did with the liquid unknown. Again, make measurements at two different solute concentrations. If the molar mass you find for your solid is less than the actual molar mass, you have an ionic solid. The ratio of the true molar mass to the value you find is equal to a quantity called the van’t Hoff factor, i. This factor can be related to the apparent percentage dissociation of the salt you are studying. For a 1:1 salt, % dissociation = (i − 1) × 100%

(3)

It turns out that Equation 3 works well for weak electrolytes like acetic acid, but is not correct for salts like NaCl, which modern theory assumes are completely ionized in aqueous solution. In your report, note all the masses observed and the temperatures of the equilibrium mixtures. Report the average molar masses you calculate and show how you obtained them. If you have an ionic solid, find i and the % dissociation as predicted by the Arrhenius theory. When you are finished with the experiment, pour the solution and the contents of the cup into the sink. Return the digital thermometer and the insulated cup to the stock room. Optional Experiment: Determine the largest possible freezing point depression for an NaCl solution. Find the composition of that solution. What phases are in equilibrium in the final mixture?



Name ___________________________________ Section ___________________________________

Experiment 19

Data and Calculations: Molar Mass Determination by Depression of the Freezing Point For this experiment you will be responsible for recording and processing your data. In the upper part of the page, note the value of each quantity that was measured, along with its units. Below all the experimental data, make the calculations called for in the procedure, and list your results. Include all the equations that you used, first in words, and then with numerical values.



Name ___________________________________ Section ___________________________________ Experiment

19

Advance Study Assignment: Determination of Molar Mass by Depression of the Freezing Point 1. A student determines the molar mass of methanol, CH3OH, by the method used in this experiment. She found that the equilibrium temperature of a mixture of ice and pure water was 0.4°C on her thermometer. When she added 10.0 g of her sample to the mixture, the temperature, after thorough stirring, fell to −5.4°C. She then poured off the solution through a screen into a beaker. The mass of the solution was 101.8 g. a. What was the freezing point depression?

______________ °C

b. What was the molality of the methanol?

______________ m

c. How much methanol was in the decanted solution?

______________ g

d. How much water was in the decanted solution?

______________ g

e. How much methanol would there be in a solution containing 1 kg of water, with methanol at the same concentration as she had in her experiment?

______________ g methanol f. What did she find to be the molar mass of methanol, assuming she made the calculation properly?

______________ g



Experiment 20 Rates of Chemical Reactions, I. The Iodination of Acetone

he rate at which a chemical reaction occurs depends on several factors: the nature of the reaction, the concentrations of the reactants, the temperature, and the presence of possible catalysts. All of these factors can markedly influence the observed rate of reaction. Some reactions at a given temperature are very slow indeed; the oxidation of gaseous hydrogen or wood at room temperature would not proceed appreciably in a century. Other reactions are essentially instantaneous; the precipitation of silver chloride when solutions containing silver ions and chloride ions are mixed and the formation of water when acidic and basic solutions are mixed are examples of extremely rapid reactions. In this experiment we will study a reaction that, in the vicinity of room temperature, proceeds at a moderate, relatively easily measured rate. For a given reaction, the rate typically increases with an increase in the concentration of any reactant. The relation between rate and concentration is a remarkably simple one in many cases, and for the reaction

T

aA + bB → cC the rate can usually be expressed by the equation rate = k(A)m(B)n

(1)

where m and n are generally, but not always, integers, 0, 1, 2, or possibly 3; (A) and (B) are the concentrations of A and B (ordinarily in moles per liter); and k is a constant, called the rate constant of the reaction, which makes the relation quantitatively correct. The numbers m and n are called the orders of the reaction with respect to A and B. If m is 1, the reaction is said to be first order with respect to the reactant A. If n is 2, the reaction is second order with respect to reactant B. The overall order is the sum of m and n. In this example the reaction would be third order overall. The rate of a reaction is also significantly dependent on the temperature at which the reaction occurs. An increase in temperature increases the rate, an often-cited rule being that a 10°C rise in temperature will double the rate. This rule is only approximately correct; nevertheless, it is clear that a rise of temperature of say 100°C could change the rate of a reaction very appreciably. As with the concentration, there is a quantitative relation between reaction rate and temperature, but here the relation is somewhat more complicated. This relation is based on the idea that to react, the reactant species must have a certain minimum amount of energy present at the time the reactants collide in the reaction step; this amount of energy, which is typically furnished by the kinetic energy of motion of the species present, is called the activation energy for the reaction. The equation relating the rate constant k to the absolute temperature T and the activation energy Ea is called the Arrhenius equation: ln k =

− Ea + constant RT

(2)

where ln k is the natural logarithm of k, and R is the gas constant (8.31 joules/mole K for Ea in joules per mole). This equation is identical in form to Equation 1 in Experiment 15. By measuring k at different temperatures we can determine graphically the activation energy for a reaction. In this experiment we will study the kinetics of the reaction between iodine and acetone: O O CH3

C

CH 3(aq) + I2(aq) → CH3

C

CH 2I(aq) + H+(aq) + I−(aq)


150

Experiment 20

Rates of Chemical Reactions, I. The Iodination of Acetone

The rate of this reaction is found to depend on the concentration of hydrogen ion in the solution as well as presumably on the concentrations of the two reactants. By Equation 1, the rate law for this reaction is rate = k(acetone)m(I2)n(H+)p

(3)

where m, n, and p are the orders of the reaction with respect to acetone, iodine, and hydrogen ion, respectively, and k is the rate constant for the reaction. The rate of this reaction can be expressed as the (small) change in the concentration of I2, Δ(I2), that occurs, divided by the time interval Δt required for the change: rate =

−Δ(I 2 ) Δt

(4)

The minus sign is to make the rate positive [as Δ(I2) is negative]. Ordinarily, since rate varies as the concentrations of the reactants according to Equation 3, in a rate study it would be necessary to measure, directly or indirectly, the concentration of each reactant as a function of time; the rate would typically vary markedly with time, decreasing to very low values as the concentration of at least one reactant becomes very low. This makes reaction rate studies relatively difficult to carry out and introduces mathematical complexities that are difficult for beginning students to understand. The iodination of acetone is a rather atypical reaction, in that it can be easily investigated experimentally. First of all, iodine has color, so that one can readily follow changes in iodine concentration visually. A second and very important characteristic of this reaction is that it turns out to be zero order in I2 concentration. This means (see Equation 3) that the rate of the reaction does not depend on (I2) at all; (I2)0 = 1, no matter what the value of (I2) is, as long as it is not itself zero. Because the rate of the reaction does not depend on (I2), we can study the rate by simply making I2 the limiting reagent present in a large excess of acetone and H+ ion. We then measure the time required for a known initial concentration of I2 to be used up completely. If both acetone and H+ are present at much higher concentrations than that of I2, their concentrations will not change appreciably during the course of the reaction, and the rate will remain, by Equation 3, effectively constant until all the iodine is gone, at which time the reaction will stop. Under such circumstances, if it takes t seconds for the color of a solution having an initial concentration of I2 equal to (I2)0 to disappear, the rate of the reaction, by Equation 4, would be rate =

− Δ ( I 2 ) ( I 2 )0 = t Δt

(5)

Although the rate of the reaction is constant during its course under the conditions we have set up, we can vary it by changing the initial concentrations of acetone and H+ ion. If, for example, we should double the initial concentration of acetone over that in Mixture 1, keeping (H+) and (I2) at the same values they had previously, then the rate of Mixture 2 would, according to Equation 3, be different from that in Mixture 1: rate 2 = k(2A)m(I2)0(H+)p

(6a)

rate 1 = k(A)m(I2)0(H+)p

(6b)

Dividing the first equation by the second, we see that the k’s cancel, as do the terms in the iodine and hydrogen ion concentrations, since they have the same values in both reactions, and we obtain simply m

rate 2 (2 A)m ⎛ 2 A ⎞ = = = 2m rate 1 ( A)m ⎝ A ⎠

(6)

Having measured both rate 2 and rate 1 by Equation 5, we can find their ratio, which must be equal to 2m. We can then solve for m either by inspection or using logarithms and so find the order of the reaction with respect to acetone. By a similar procedure we can measure the order of the reaction with respect to H+ ion concentration and also confirm the fact that the reaction is zero order with respect to I2. Having found the order with respect to each reactant, we can then evaluate k, the rate constant for the reaction. The determination of the orders m and p, the confirmation of the fact that n, the order with respect to I2, equals zero, and the evaluation of the rate constant k for the reaction at room temperature comprise your


Experiment 20


151

assignment in this experiment. You will be furnished with standard solutions of acetone, iodine, and hydrogen ion, and with the composition of one solution that will give a reasonable rate. The rest of the planning and the execution of the experiment will be your responsibility. An optional part of the experiment is to study the rate of this reaction at different temperatures to find its activation energy. The general procedure here would be to study the rate of reaction in one of the mixtures at room temperature and at two other temperatures, one above and one below room temperature. Knowing the rates, and hence the k’s, at the three temperatures, you can then find Ea, the energy of activation for the reaction, by plotting ln k vs. 1/T. The slope of the resultant straight line, by Equation 2, must be −Ea/R.

Experimental Procedure Select two regular (18 × 150 mm) test tubes; when filled with distilled water, they should appear to have identical color when you view them down the tubes against a white background. Draw 50 mL of each of the following solutions into clean, dry, 100-mL beakers, one solution to a beaker: 4 M acetone, 1 M HCl, and 0.005 M I2. Cover each beaker with a watch glass. With your graduated cylinder, measure out 10.0 mL of the 4 M acetone solution and pour it into a clean 125-mL Erlenmeyer flask. Then measure out 10.0 mL 1 M HCl and add that to the acetone in the flask. Add 20.0 mL distilled H2O to the flask. Drain the graduated cylinder, shaking out any excess water, and then use the cylinder to measure out 10.0 mL 0.005 M I2 solution. Be careful not to spill the iodine solution on your hands or clothes. Noting the time on your wristwatch or the wall clock to 1 second, pour the iodine solution into the Erlenmeyer flask and quickly swirl the flask to mix the reagents thoroughly. The reaction mixture will appear yellow because of the presence of the iodine, and the color will fade slowly as the iodine reacts with the acetone. Fill one of the test tubes 3/4 full with the reaction mixture, and fill the other test tube to the same depth with distilled water. Look down the test tubes toward a well-lit piece of white paper, and note the time the color of the iodine just disappears. Measure the temperature of the mixture in the test tube. Repeat the experiment, using as a reference the reacted solution instead of distilled water. The amount of time required in the two runs should agree within about 20 seconds. The rate of the reaction equals the initial concentration of I2 in the reaction mixture divided by the elapsed time. Since the reaction is zero order in I2, and since both acetone and H+ ion are present in great excess, the rate is constant throughout the reaction and the concentrations of both acetone and H+ remain essentially at their initial values in the reaction mixture. Having found the reaction rate for one composition of the system, think for a moment about what changes in composition you might make to decrease the time and hence increase the rate of reaction. In particular, how could you change the composition in such a way as to allow you to determine how the rate depends upon acetone concentration? If it is not clear how to proceed, reread the discussion preceding Equation 6. In your new mixture you should keep the total volume at 50 mL, and be sure that the concentrations of H+ and I2 are the same as in the first experiment. Carry out the reaction twice with your new mixture; the times should not differ by more than about 15 seconds. The temperature should be kept within about a degree of that in the initial run. Calculate the rate of the reaction. Compare it with that for the first mixture, and then calculate the order of the reaction with respect to acetone, using a relation similar to Equation 6. First, write an equation like 6a for the second reaction mixture, substituting in the values for the rate as obtained by Equation 5 and the initial concentration of acetone, I2, and H+ in the reaction mixture. Then write an equation like 6b for the first reaction mixture, using the observed rate and the initial concentrations in that mixture. Obtain an equation like 6 by dividing Equation 6a by Equation 6b. Solve Equation 6 for the order m of the reaction with respect to acetone. Again change the composition of the initial reaction mixture so that this time a measurement of the reaction rate will give you information about the order of the reaction with respect to H+. Repeat the experiment with this mixture to establish the time of reaction to within 15 seconds, again making sure that the temperature is within about a degree of that observed previously. From the rate you determine for this mixture find p, the order of the reaction with respect to H+.


152

Experiment 20


Finally, change the reaction mixture composition in such a way as to allow you to show that the order of the reaction with respect to I2 is zero. Measure the rate of the reaction twice, and calculate n, the order with respect to I2. Having found the order of the reaction for each species on which the rate depends, evaluate k, the rate constant for the reaction, from the rate and concentration data in each of the mixtures you studied. If the temperatures at which the reactions were run are all equal to within a degree or two, k should be about the same for each mixture. Calculate the mean value of k and the standard deviation for the set of values. (See Appendix VIII.) Optional As a final reaction, make up a mixture using reactant volumes that you did not use in any previous experiments. Using Equation 3, the values of concentrations in the mixtures, the orders, and the rate constant you calculated from your experimental data, predict how long it will take for the I2 color to disappear from your mixture. Measure the time for the reaction and compare it with your prediction. If time permits, select one of the reaction mixtures you have already used that gave a convenient time, and use that mixture to measure the rate of reaction at about 10°C and at about 40°C. From the two rates you find, plus the rate at room temperature, calculate the energy of activation for the reaction, using Equation 2. DISPOSAL OF REACTION PRODUCTS.

Dispose of your solutions from this experiment as directed

by your laboratory supervisor.


Name ____________________________________

Section ________________________________

Experiment 20

Data and Calculations: The Iodination of Acetone A. Reaction Rate Data Volume in mL 4.0 M Acetone

Volume in mL 1.0 M HCl

Volume in mL 0.0050 M I2

Volume in mL H2O

I

10 _________

10 _________

10 _________

20 _________

_________

_________ _________

II

_________

_________

_________

_________

_________

_________ _________

III

_________

_________

_________

_________

_________

_________ _________

IV

_________

_________

_________

_________

_________

_________ _________

Mixture

Time for Reaction in sec 1st Run 2nd Run

Temp. in °C

B. Determination of Reaction Orders with Respect to Acetone, H+ Ion, and I2 rate = k(acetone)m(I2)n(H+)p

(3)

Calculate the initial concentrations of acetone, H+ ion, and I2 in each of the mixtures you studied. Use Equation 5 to find the rate of each reaction. (I 2 )0 avg. time

(Acetone)

(H+)

(I2)0

I

0.80 M ______________

0.20 M ______________

0.0010 M ______________

______________

II

______________

______________

______________

______________

III

______________

______________

______________

______________

IV

______________

______________

______________

______________

Mixture

Rate =

Substituting the initial concentrations and the rate from the table above, write Equation 3 as it would apply to Reaction Mixture II: Rate II = Now write Equation 3 for Reaction Mixture I, substituting concentrations and the calculated rate from the table: Rate I = (continued on following page)


154

Experiment 20


Divide the equation for Mixture II by the equation for Mixture I; the resulting equation should have the ratio of Rate II to Rate I on the left side, and a ratio of acetone concentrations raised to the power m on the right. It should be similar in appearance to Equation 6. Put the resulting equation below: Rate II = Rate I The only unknown in the equation is m. Solve for m.

m = _____________

Now write Equation 3 as it would apply to Reaction Mixture III and as it would apply to Reaction Mixture IV: Rate III = Rate IV = Using the ratios of the rates of Mixtures III and IV to those of Mixtures II or I, find the orders of the reaction with respect to H+ ion and I2: Rate III

= p = _____________

Rate ________ Rate IV

= n = _____________

Rate ________

C. Determination of the Rate Constant k Given the values of m, p, and n as determined in Part B, calculate the rate constant k for each mixture by simply substituting those orders, the initial concentrations, and the observed rate from the table into Equation 3. Mixture k

I _____________

II _____________

III _____________

IV _____________

Average _____________

Standard deviation in k _____________

D. Prediction of Reaction Rate

Optional

Reaction mixture Volume Volume Volume Volume in mL in mL in mL in mL 4.0 M acetone ____________ 1.0 M HCl ____________ 0.0050 M I2 ____________ H2O ____________ Initial concentrations (acetone) _____________ M

(H+) _____________ M

(I2)0 _____________ M

Predicted rate _____________ (Eq. 3) (continued on following page) Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Experiment 20


155

Predicted time for reaction _____________ sec (Eq. 5) Observed time for reaction _____________ sec

E. Determination of Energy of Activation

Optional

Reaction mixture used _____________ (same for all temperatures) Time for reaction at about 10°C ___________ sec

Temperature ___________ °C; ___________ K

Time for reaction at about 40°C ___________ sec

Temperature ___________ °C; ___________ K

Time for reaction at room temp ___________ sec

Temperature ___________ °C; ___________ K

Calculate the rate constant at each temperature from your data, following the procedure in Part C.

∼10°C

Rate _____________

k _____________

ln k _____________

1 Τ (Κ ) _____________

∼40°C

_____________

_____________

_____________

_____________

Room temp

_____________

_____________

_____________

_____________

Plot ln k vs. 1/T using Excel or the graph paper on the following page. Find the slope of the best straight line through the points. (If you need to, see Appendix V.)

Slope = _____________ By Equation 2:

Ea = −8.31 × slope Ea = _____________ joules


Name ____________________________________

Section ________________________________

Experiment 20

The Iodination of Acetone (Data and Calculations)


Name ____________________________________

Section ________________________________

Experiment 20

Advance Study Assignment: The Iodination of Acetone 1. In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture: 10 mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2 + 20 mL H2O a. How many moles of acetone were in the reaction mixture? Recall that, for a component A, no. moles A = MA × V, where MA is the molarity of A and V is the volume in liters of the solution of A that was used.

___________ moles acetone b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50 mL, 0.050 L, and the number of moles of acetone was found in Part (a). Again, no. moles A MA = V of soln. in liters ___________ M acetone c. How could you double the molarity of the acetone in the reaction mixture, keeping the total volume at 50 mL and keeping the same concentrations of H+ ion and I2 as in the original mixture? 2. Using the reaction mixture in Problem 1, a student found that it took 230 seconds for the color of the I2 to disappear. a. What was the rate of the reaction? Hint: First find the initial concentration of I2 in the reaction mixture, (I2)0. Then use Equation 5. rate = ___________ b. Given the rate from Part (a), and the initial concentrations of acetone, H+ ion, and I2 in the reaction mixture, write Equation 3 as it would apply to the mixture. rate = c. What are the unknowns that remain in the equation in Part (b)? ___________ ___________ ___________ ___________ (continued on following page)


158

Experiment 20


3. A second reaction mixture was made up in the following way: 10 mL 4.0 M acetone + 20 mL 1.0 M HCl + 10 mL 0.0050 M I2 + 10 mL H2O a. What were the initial concentrations of acetone, H+ ion, and I2 in the reaction mixture? (acetone) ___________ M; (H+) ___________ M; (I2)0 ___________ M b. It took 120 seconds for the I2 color to disappear from the reaction mixture when it occurred at the same temperature as the reaction in Problem 2. What was the rate of the reaction? ___________

Write Equation 3 as it would apply to the second reaction mixture: rate = c. Divide the equation in Part (b) by the equation in Problem 2b. The resulting equation should have the ratio of the two rates on the left side and a ratio of acetone concentrations raised to the m power on the right. Write the resulting equation and solve for the value of m, the order of the reaction with respect to acetone. (Round off the value of m to the nearest integer.)

m = ___________ 4. A third reaction mixture was made up in the following way: 10 mL 4.0 M acetone + 20 mL 1.0 M HCl + 5 mL 0.0050 M I2 + 15 mL H2O If the reaction is zero order in I2, how long would it take for the I2 color to disappear at the temperature of the reaction mixture in Problem 3? ___________ seconds


Experiment 21 Rates of Chemical Reactions, II. A Clock Reaction

n the previous experiment we discussed the factors that influence the rate of a chemical reaction and presented the terminology used in quantitative relations in studies of the kinetics of chemical reactions. That material is also pertinent to this experiment and should be studied before you proceed further. This experiment involves the study of the rate properties, or chemical kinetics, of the following reaction between iodide ion and bromate ion under acidic conditions:

I

6 I−(aq) + BrO3−(aq) + 6 H+(aq) → 3 I2(aq) + Br−(aq) + 3 H2O

(1)

This reaction proceeds reasonably slowly at room temperature, its rate depending on the concentrations of the I−, BrO3−, and H+ ions according to the rate law discussed in the previous experiment. For this reaction the rate law takes the form rate = k(I−)m(BrO3−)n(H+)p

(2)

One of the main purposes of this experiment will be to evaluate the rate constant k and the reaction orders m, n, and p for this reaction. We will also investigate the manner in which the reaction rate depends on temperature, and will evaluate the activation energy of the reaction, Ea. Our method for measuring the rate of the reaction involves what is frequently called a clock reaction. In addition to Reaction 1, whose kinetics we will study, the following reaction will also be made to occur simultaneously in the reaction flask: I2(aq) + 2 S2O32−(aq) → 2 I−(aq) + S4O62−(aq)

(3)

As compared with Equation 1 this reaction is essentially instantaneous. The I2 produced in (1) reacts completely with the thiosulfate, S2O32−, ion present in the solution, so that until all the thiosulfate ion has reacted, the concentration of I2 is effectively zero. As soon as the S2O32− is gone from the system, the I2 produced by (1) remains in the solution and its concentration begins to increase. The presence of I2 is made strikingly apparent by a starch indicator that is added to the reaction mixture, since I2 even in small concentrations reacts with starch solution to produce a deep blue color. By carrying out Reaction 1 in the presence of S2O32− and a starch indicator, we introduce a “clock” into the system. Our clock tells us when a given amount of BrO3− ion has reacted (1/6 mole BrO3− per mole S2O32−), which is just what we need to know, since the rate of reaction can be expressed in terms of the time it takes for a particular amount of BrO3− to be used up. In all our reactions, the amount of BrO3− that reacts in the time we measure will be constant and small as compared to the amounts of any of the other reactants. This means that the concentrations of all reactants will be essentially constant in Equation 2, and hence so will the rate during each reaction. In our experiment we will carry out the reaction between BrO3−, I−, and H+ ions under different concentration conditions. Measured amounts of each of these ions in aqueous solution will be mixed in the presence of a constant, small amount of S2O32−. The time it takes for each mixture to turn blue will be measured. The time obtained for each reaction will be inversely proportional to its rate. By changing the concentration of one reactant and keeping the other concentrations constant, we can investigate how the rate of the reaction varies with the concentration of a particular reactant. Once we know the order for each reactant we can determine the rate constant for the reaction. In the last part of the experiment we will investigate how the rate of the reaction depends on temperature. You will recall that in general the rate increases sharply with temperature. By measuring how the rate varies


160

Experiment 21

Rates of Chemical Reactions, II. A Clock Reaction

with temperature we can determine the activation energy, Ea, for the reaction by making use of the Arrhenius equation: ln k = −

Ea + constant RT

(4)

In this equation, k is the rate constant at the Kelvin temperature T, Ea is the activation energy, and R is the gas constant. By plotting ln k against 1/T we should obtain, by Equation 4, a straight line whose slope equals −Ea/R. From the slope of that line we can easily calculate the activation energy.

Experimental Procedure A. Dependence of Reaction Rate on Concentration In Table 21.1 we have summarized the reagent volumes to be used in carrying out the several reactions whose rates we need to know to find the general rate law for Reaction 1. First, measure out 100 mL of each of the listed reagents (except H2O) into clean, labeled flasks or beakers. Use these reagents in your reaction mixtures.

Table 21.1 Reaction Mixtures at Room Temperature (Reagent volumes in mL) Reaction Mixture 1 2 3 4 5

Reaction Flask I (250 mL)

Reaction Flask II (125 mL)

0.010 M KI

0.0010 M Na2S2O3

H2O

0.040 M KBrO3

0.10 M HCl

10 20 10 10 8

10 10 10 10 10

10 0 0 0 12

10 10 20 10 5

10 10 10 20 15

The actual procedure for each reaction mixture will be much the same, and we will describe it now for Reaction Mixture 1. Since there are several reagents to mix, and since we don’t want the reaction to start until we are ready, we will put some of the reagents into one flask and the rest into another, selecting them so that no reaction occurs until the contents of the two flasks are mixed. Using a 10-mL graduated cylinder to measure volumes, measure out 10 mL 0.010 M KI, 10 mL 0.0010 M Na2S2O3, and 10 mL distilled water into a 250-mL Erlenmeyer flask (Reaction Flask I). Then measure out 10 mL 0.040 M KBrO3 and 10 mL 0.10 M HCl into a 125-mL Erlenmeyer flask (Reaction Flask II). To Flask II add three or four drops of starch indicator solution. Pour the contents of Reaction Flask II into Reaction Flask I and swirl the solutions to mix them thoroughly. Note the time at which the solutions were mixed. Continue swirling the solution. It should turn blue in less than 2 minutes. Note the time at the instant that the blue color appears. Record the temperature of the blue solution to 0.2°C. Repeat the procedure with the other mixtures in Table 21.1. Don’t forget to add the indicator before mixing the solutions in the two flasks. The reaction flasks should be rinsed with distilled water between runs. When measuring out reagents, rinse the graduated cylinder with distilled water after you have added the reagents to Reaction Flask I, and before you measure out the reagents for Reaction Flask II. Try to keep the temperature just about the same in all the runs. Repeat any experiments that did not appear to proceed properly.


Experiment 21


161

B. Dependence of Reaction Rate on Temperature In this part of the experiment, the reaction will be carried out at several different temperatures, using Reaction Mixture 1 in all cases. The temperatures we will use will be about 20°C, 40°C, 10°C, and 0°C. We will take the time at about 20°C to be that for Reaction Mixture 1 as determined at room temperature. To determine the time at 40°C proceed as follows. Make up Reaction Mixture 1 as you did in Part A, including the indicator. However, instead of mixing the solutions in the two flasks at room temperature, put the flasks into water at 40°C, drawn from the hot-water tap into one or more large beakers. Check to see that the water is indeed at about 40°C, and leave the flasks in the water for several minutes to bring them to the proper temperature. Then mix the two solutions, noting the time of mixing. Continue swirling the reaction flask in the warm water. When the color change occurs, note the time and the temperature of the solution in the flask. Repeat the experiment at about 10°C, cooling all the reactants in water at that temperature before starting the reaction. Record the time required for the color to change and the final temperature of the reaction mixture. Repeat once again at about 0°C, this time using an ice-water bath to cool the reactants.

C. Dependence of the Reaction Rate on the Presence of Catalyst Optional Some ions have a pronounced catalytic effect on the rates of many reactions in water solution. Observe the effect on this reaction by once again making up Reaction Mixture 1. Before mixing, add one drop 0.5 M (NH4)2MoO4, ammonium molybdate, and a few drops of starch indicator to Reaction Flask II. Swirl the flask to mix the catalyst thoroughly. Then mix the solutions, noting the time required for the color to change. DISPOSAL OF REACTION PRODUCTS. The reaction products in this experiment are very dilute and may be poured into the sink as you complete each part of the experiment if so directed by your instructor.



Name ____________________________________

Section _________________________________

Experiment 21

Data and Calculations: Rates of Chemical Reactions, II. A Clock Reaction A. Dependence of Reaction Rate on Concentration Reaction: 6 I−(aq) + BrO3−(aq) + 6 H+(aq) → 3 I2(aq) + Br−(aq) + 3 H2O

(1)

Δ(BrO3− ) (2) t In all the reaction mixtures used in this experiment, the color change occurred when a constant predetermined number of moles of BrO3− had been used up by the reaction. The color “clock” allows you to measure the time required for this fixed number of moles of BrO3− to react. The rate of each reaction is determined by the time t required for the color to change; since in Equation 2 the change in concentration of BrO3− ion, Δ(BrO3−), is the same in each mixture, the relative rate of each reaction is inversely proportional to the time t. Since we are mainly concerned with relative rather than absolute rate, we will for convenience take all relative rates as being equal to 1000/t. Fill in the following table, first calculating the relative reaction rate for each mixture. rate = k(I−)m(BrO3−)n(H+)p = −

Time t (sec) Reaction for Color Mixture to Change

Reactant Concentrations in Reacting Mixture (M)

Relative Rate of Reaction 1000/t

(I-)

(BrO3-)

(H+)

Temp. in (°C)

1

__________

__________

0.0020 __________

__________

__________

__________

2

__________

__________

__________

__________

__________

__________

3

__________

__________

__________

__________

__________

__________

4

__________

__________

__________

__________

__________

__________

5

__________

__________

__________

__________

__________

__________

The reactant concentrations in the reaction mixture are not those of the stock solutions, since the reagents were diluted by the other solutions. The final volume of the reaction mixture is 50 mL in all cases. Since the number of moles of reactant does not change on dilution we can say, for example, for I− ion, that no. moles I− = (I−)stock × Vstock = (I−)mixture × Vmixture For Reaction Mixture 1, (I−)stock = 0.010 M,

Vstock = 10 mL,

Vmixture = 50 mL

Therefore, 0.010 M × 10 mL = 0.0020 M 50 mL Calculate the rest of the concentrations in the table using the same approach. (I − )mixture =

Determination of the Orders of the Reaction Given the data in the table, the problem is to find the order for each reactant and the rate constant for the reaction. Since we are dealing with relative rates, we can modify Equation 2 to read as follows: relative rate = k′(I−)m(BrO3−)n(H+)p

(5) (continued on the following page)


164

Experiment 21


We need to determine the relative rate constant k′ and the orders m, n, and p in such a way as to be consistent with the data in the table. The solution to this problem is quite simple, once you make a few observations on the reaction mixtures. Each mixture (2 to 4) differs from Reaction Mixture 1 in the concentration of only one species (see table). This means that for any pair of mixtures that includes Reaction Mixture 1, there is only one concentration that changes. From the ratio of the relative rates for such a pair of mixtures we can find the order for the reactant whose concentration was changed. Proceed as follows. Write Equation 5 below for Reaction Mixtures 1 and 2, substituting the relative rates and the concentrations of I−, BrO3−, and H+ ions from the table you have just completed. Relative Rate 1 = ____________ = k′(

)m(

)n(

)p

Relative Rate 2 = ____________ = k′(

)m(

)n(

)p

Divide the first equation by the second, noting that nearly all the terms cancel out. The result is simply Relative Rate 1 = Relative Rate 2 If you have done this properly, you will have an equation involving only m as an unknown. Solve this equation for m, the order of the reaction with respect to I− ion. m = ____________ (nearest integer) Applying the same approach to Reaction Mixtures 1 and 3, find the value of n, the order of the reaction with respect to BrO3− ion. Relative Rate 1 = ____________ = k′(

)m(

)n(

)p

Relative Rate 3 = ____________ = k′(

)m(

)n(

)p

Dividing one equation by the other: =

n = ____________

Now that you have the idea, apply the method once again, this time to Reaction Mixtures 1 and 4, and find p, the order with respect to H+ ion. Relative Rate 4 = k′(

)m(

)n(

)p

Dividing the equation for Relative Rate 1 by that for Relative Rate 4, we get =

p = ____________

Having found m, n, and p (nearest integers), the relative rate constant, k′, can be calculated by substitution of m, n, p, and the known rates and reactant concentrations into Equation 5. Evaluate k′ for Reaction Mixtures 1 to 4. Reaction 1 2 3 k′ ____________ ____________ ____________ Standard deviation in k'____________ (See Appendix VIII)

4 ____________

k′ave ____________

Why should k′ have nearly the same value for each of the above reactions? Using k′ave in Equation 5, predict the relative rate and time, tpred, for Reaction Mixture 5. Use the concentrations in the table. Relative ratepred ____________ tpred ____________ tobs ____________ (continued on the following page) Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Experiment 21


165

B. Effect of Temperature on Reaction Rate: The Activation Energy To find the activation energy for the reaction it will be helpful to complete the following table. The dependence of the rate constant, k′, for a reaction is given by Equation 4: ln k ′ = −

Ea + constant RT

(4)

Since the reactions at the different temperatures all involve the same reactant concentrations, the rate constants, k′, for two different mixtures will have the same ratio as the reaction rates themselves for the two mixtures. This means that in the calculation of Ea, we can use the observed relative rates instead of rate constants. Proceeding as before, calculate the relative rates of reaction in each of the mixtures and enter these values in (c). Take the ln rate for each mixture and enter these values in (d). To set up the terms in 1/T, fill in (b), (e), and (f) in the table. Approximate Temperature in °C

(a) Time t in seconds for color to appear

20 ________

40 ________

10 ________

0 ________

(b) Temperature of the reaction mixture in °C

________

________

________

________

(c) Relative rate = 1000/t

________

________

________

________

(d) ln of relative rate

________

________

________

________

(e) Temperature T in K

________

________

________

________

(f) 1/T, K−1

________

________

________

________

To evaluate Ea, make a graph of ln relative rate vs. 1/T, using Excel or the graph paper provided. (See Appendix V.) Find the slope of the line obtained by drawing the best straight line through the experimental points. Slope = ____________ The slope of the line equals −Ea/R, where R = (8.31 joules/mole K) if Ea is to be in joules per mole. Find the activation energy, Ea, for the reaction. Ea = ____________ joules/mole

C. Effect of a Catalyst on Reaction Rate

Time for color to appear (seconds)

Reaction 1 ____________

Optional

Catalyzed Reaction 1 ____________

Would you expect the activation energy, Ea, for the catalyzed reaction to be greater than, less than, or equal to the activation energy for the uncatalyzed reaction? Why?


Name ____________________________________

Section _________________________________

Experiment 21

Rates of Chemical Reactions, II. A Clock Reaction (Data and Calculations)


Name ____________________________________

Section _________________________________

Experiment 21

Advance Study Assignment: Rates of Chemical Reactions, II. A Clock Reaction 1. A student studied the clock reaction described in this experiment. She set up Reaction Mixture 4 by mixing 10 mL 0.010 M KI, 10 mL 0.001 M Na2S2O3, 10 mL 0.040 M KBrO3, and 20 mL 0.10 M HCl using the procedure given. It took about 21 seconds for the color to turn blue. a. She found the concentrations of each reactant in the reacting mixture by realizing that the number of moles of each reactant did not change when that reactant was mixed with the others, but that its concentration did. For any reactant A, no. moles A = MA stock × Vstock = MA mixture × Vmixture The volume of the mixture was 50 mL. Revising the above equation, she obtained Vstock ( mL) 50 mL

MA mixture = MA stock ×

Find the concentrations of each reactant using the equation above. (I−) = ____________ M; (BrO3−) = ____________ M; (H+) = ____________ M b. What was the relative rate of the reaction (1000/t)? ____________ c. Knowing the relative rate of reaction for Mixture 4 and the concentrations of I−, BrO3−, and H+ in that mixture, she was able to set up Equation 5 for the relative rate of the reaction. The only quantities that remained unknown were k′, m, n, and p. Set up Equation 5 as she did, presuming she did it properly. Equation 5 is on page 163. 2. For Reaction Mixture 1 the student found that 85 seconds were required. On dividing Equation 5 for Reaction Mixture 1 by Equation 5 for Reaction Mixture 4, and after canceling out the common terms [k′ and terms in (I−) and (BrO3−)], she got the following equation: p

11.8 ⎛ 0.020 ⎞ 1 = =⎛ ⎞ 48 ⎝ 0.040 ⎠ ⎝ 2⎠

p

Recognizing that 11.8/48 is about equal to 14 , she obtained an approximate value for m. What was that value? m = ____________ By taking logarithms of both sides of the equation, she got an exact value for m. What was that value? m = ____________ Since orders of reactions are often integers, she rounded her value of m to the nearest integer, and reported that value as the order of the reaction with respect to H+.



Experiment 22 Properties of Systems in Chemical Equilibrium—Le Châtelier’s Principle

hen working in the laboratory, one often makes observations that at first sight are surprising and hard to explain. One might add a reagent to a solution and obtain a precipitate. Addition of more of that reagent to the precipitate causes it to dissolve. A violet solution turns yellow on addition of a reagent. Subsequent addition of another reagent brings back first a green solution and then the original violet one. Clearly, chemical reactions are occurring, but how and why they behave as they do is not at once obvious. In this experiment we will examine and attempt to explain several observations of the sort we have just mentioned. Central to our explanation will be recognition of the fact that chemical systems tend to exist in a state of equilibrium. If one disturbs the equilibrium in one way or another, the reaction may shift to the left or right, producing the kinds of effects we have mentioned. If one can understand the principles governing the equilibrium system, it is often possible to see how one might disturb the system, such as by adding a particular reagent or heat, and so cause it to change in a desirable way. Before proceeding to specific examples, let us examine the situation in a general way, noting the key principle that allows us to make a system in equilibrium behave as we wish. Consider the reaction

W

A(aq)

B(aq) + C(aq)

(1)

where A, B, and C are molecules or ions in solution. If we have a mixture of these species in equilibrium, it turns out that their concentrations are not completely unrelated. Rather, there is a condition that those concentrations must meet, namely that [B] × [C] = Kc [A]

(2)

where Kc is a constant, called the equilibrium constant for the reaction. For a given reaction at any given temperature, Kc has a particular value. When we say that Kc has a particular value, we mean just that. For example, we might find that, for a given solution in which Reaction 1 can occur, when we substitute the equilibrium values for the molarities of A, B, and C into Equation 2, we get a value of 10 for Kc. Now, suppose we add more of species A to that solution. What will happen? Remember, Kc can’t change. If we substitute the new higher molarity of A into Equation 2 we get a value that is smaller than Kc. This means that the system is not in equilibrium, and must change in some way to get back to equilibrium. How can it do this? It can do this by shifting to the right, producing more B and C and using up some A. It must do this, and will, until the molarities of C, B, and A reach values that, on substitution into Equation 2, equal 10. At that point the system is once again in equilibrium. In the new equilibrium state, [B] and [C] are greater than they were initially, and [A] is larger than its initial value but smaller than if there had been no forced shift to the right. The conclusion you should reach on reading the last paragraph is that one can always cause a reaction to shift to the right by increasing the concentration of a reactant. An increase in concentration of a product will force a shift to the left. By a similar argument we find that a decrease in reactant concentration causes a shift to the left; a decrease in product concentration produces a shift to the right. This is all true because Kc does not change (unless you change the temperature). The changes in concentration that one can produce by adding particular reagents may be simply enormous, so the shifts in the equilibrium system may also be enormous. Much of the mystery of chemical behavior disappears once you understand this idea.


170

Experiment 22

Properties of Systems in Chemical Equilibrium—Le Châtelier’s Principle

Another way one might disturb an equilibrium system is by changing its temperature. When this happens, the value of Kc changes. It turns out that the change in Kc depends upon the enthalpy change, ΔH, for the reaction. If ΔH is positive, greater than zero (an “endothermic reaction”), Kc increases with increasing T. If ΔH is negative (an “exothermic” reaction), Kc decreases with an increase in T. Let us return to our original equilibrium between A, B, and C, where Kc equals 10. Let us assume that ΔH for Reaction 1 is −40 kJ. If we raise the temperature, Kc will go down (ΔH < 0), say to a value of 1. This means that the system will no longer be in equilibrium. Substitution of the initial values of [A], [B], and [C] into Equation 2 produces a value that is too big, 10 instead of 1. How can the system change itself to regain equilibrium? It must of necessity shift to the left, lowering [B] and [C] and raising [A]. This will make the expression in Equation 2 smaller. The shift will continue until the concentrations of A, B, and C, on substitution into Equation 2, give the expression a value of 1. From the discussion in the previous paragraph, you should be able to conclude that an equilibrium system will shift to the left on being heated if the reaction is exothermic (ΔH < 0, Kc decreases with temperature). It will shift to the right if the reaction is endothermic (ΔH > 0, Kc increases with temperature). Again, since we can change temperatures very markedly, we can shift equilibria a long, long way. An endothermic reaction that at 25°C has an equilibrium state that consists mainly of reactants might, at 1000°C, exist almost completely as products. The effects of concentration and temperature on systems in chemical equilibrium are often summarized by Le Châtelier’s principle. The principle states that: If you attempt to change a system in chemical equilibrium, it will react in such a way as to counteract the change you attempted. If you think about the principle for a while, you will see that it predicts the same kind of behavior as we did by using the properties of Kc. Increasing the concentration of a reactant will, by the principle, cause a change that decreases that concentration; that change must be a shift to the right. Increasing the temperature of a reaction mixture will cause a change that tends to absorb heat; that change must be a shift in the endothermic direction. The principle is an interesting one, but does require more careful reasoning in some cases than the more direct approach we employed. For the most part we will find it more useful to base our arguments on the properties of Kc. In working with aqueous systems, the most important equilibrium is often that which involves the dissociation of water into H+ and OH− ions: H2O

H+(aq) + OH−(aq) Kc = [H+] [OH−] = 1 × 10−14

(3)

In this reaction the concentration of water is very high and is essentially constant at about 55 M; it is incorporated into Kc. The value of Kc is very small, which means that in any water system the product of [H+] and [OH−] must be very small. In pure water, [ H+] equals [OH−] equals 1 × 10−7 M. Although the product, [H+] × [OH−] is small, that does not mean that both concentrations are necessarily small. If, for example, we dissolve HCl in water, the HCl in the solution will dissociate completely to H+ and Cl− ions; in 1 M HCl, [H+] will become 1 M, and there is nothing that Reaction 3 can do about changing that concentration appreciably. Rather, Reaction 3 must occur in such a direction as to maintain equilibrium. It does this by lowering [OH−] by reaction to the left; this uses up a little bit of H+ ion and drives [OH−] to the value it must have when [H+] is 1 M, namely, 1 × 10−14 M. In 1 M HCl, [OH−] is a factor of ten million smaller than it is in water. This makes the properties of 1 M HCl quite different from those of water, particularly where H+ and OH− ions are involved. If we take 1 M HCl and add a solution of NaOH to it, an interesting situation develops. Like HCl, NaOH is completely dissociated in solution, so in 1 M NaOH, [OH−] is equal to 1 M. If we add 1 M NaOH to 1 M HCl, we will initially raise [OH−] ions way above 1 × 10−14 M. However, Reaction 3 cannot be in equilibrium when both [H+] and [OH−] are high; reaction must occur to re-establish equilibrium. The added OH− ions react with H+ ions to form H2O, decreasing both concentrations until equilibrium is established. If only a small amount of OH− ion is added, it will essentially all be used up; [H+] will remain high, and [OH−] will still be very small, but somewhat larger than 10−14 M. If we add OH− ion until the amount added equals in moles the amount of H+ originally present, then Reaction 3 will go to the left until [H+] equals [OH−] equals 1 × 10−7 M, and both concentrations will be very small. Further addition of OH− ion will raise [OH−] to much higher


Experiment 22


171

values, easily as high as 1 M. In such a solution, [H+] would be very low, 1 × 10−14 M. So, in aqueous solution, depending on the solutes present, we can have [H+] and [OH−] range from about 1 M to 10−14 M, or 14 orders of magnitude. This will have a tremendous effect on any other equilibrium system in which [H+] or [OH−] ions are reactants. Similar situations arise in other equilibrium systems in which the concentration of a reactant or product can be changed significantly by adding a particular reagent. In many equilibrium systems, several equilibria are present simultaneously. For example, in aqueous solution, Reaction 3 must always be in equilibrium. There may, in addition, be equilibria between the solutes in the aqueous solution. Some examples are those in Reactions 4, 5, 7, 8, 9, and 10 in the Experimental Procedure section. In some of those reactions, H+ and OH− ions appear; in others, they do not. In Reaction 4, for example, H+ ion is a product. The molarity of H+ in Reaction 4 is not determined by the indicator HMV, since it is only present in a tiny amount. Reaction 4 will have an equilibrium state that is fixed by the state of Reaction 3, which as we have seen depends markedly on the presence of solutes such as HCl or NaOH. Reactions 8 and 9 can similarly be controlled by Reaction 3. Reactions 5 and 7, which do not involve H+ or OH− ions, are not dependent on Reaction 3 for their equilibrium state. Reaction 10 is sensitive to NH3 concentration and can be driven far to the right by addition of a reagent such as 6 M NH3.

Experimental Procedure In this experiment we will work with several equilibrium systems, each of which is similar to the A-B-C system we discussed. We will alter these systems in various ways, forcing shifts to the right and left by changing concentrations and temperature. You will be asked to interpret your observations in terms of the principles we have presented.

A. Acid-Base Indicators There is a large group of chemical substances, called acid-base indicators, which change color in solution when [H+] changes. A typical substance of this sort is called methyl violet, which we will give the formula HMV. In solution HMV dissociates as follows: HMV(aq) yellow

H+(aq) + MV−(aq) violet

(4)

HMV has an intense yellow color, while the anion MV− is violet. The color of the indicator in solution depends very strongly on [H+]. Step 1 Add about 5 mL of distilled water to a regular (18 × 150 mm) test tube. Add a few drops of

methyl violet indicator. Report the color of the solution on the Data page. Step 2 How could you force the equilibrium system to go to the other form (color)? Select a reagent

that should do this and add it to the solution, drop by drop, until the color change is complete. If your reagent works, write its formula on the Data page. If it doesn’t, try another until you find one that does. Work with 6 M reagents if they are available. Step 3 Equilibrium systems are reversible. That is, the reaction can be driven to the left and right

many times by changing the conditions in the system. How can you force the system in Step 2 to revert to its original color? Select a reagent that should do this and add it drop by drop until the color has become the original one. Again, if your first choice was incorrect, try another reagent. On the Data page write the formula of the reagent that was effective. Answer all the questions for Part A before going on to Part B.


172

Experiment 22


B. Solubility Equilibrium; Finding a Value for Ksp Many ionic substances have limited water solubility. A typical example is PbCl2, which dissolves to some extent in water according to the reaction PbCl2(s)

Pb2+(aq) + 2 Cl−(aq)

(5)

The equilibrium constant for this reaction takes the form Kc = [Pb2+] × [Cl−]2 = Ksp

(6)

The PbCl2 does not enter into the expression because it is a solid, and so has a constant effect on the system, independent of its amount. The equilibrium constant for a solubility equilibrium is called the solubility product, and is given the symbol Ksp. For the equilibrium in Reaction 5 to exist, there must be some solid PbCl2 present in the system. If there is no solid, there is no equilibrium; Equation 6 is not obeyed, and [Pb2+] × [Cl−]2 must be less than the value of Ksp. If the solid is present, even in a tiny amount, then the values of [Pb2+] and [Cl−] are subject to Equation 6. Step 1 Set up a hot-water bath, using a 400-mL beaker half full of water. Start heating the water

with a burner while proceeding with Step 2. Step 2 To a regular test tube add 5.0 mL 0.30 M Pb(NO3)2. In this solution [Pb2+] equals 0.30 M.

Add 5.0 mL 0.30 M HCl to a 10-mL graduated cylinder. In this solution [Cl−] equals 0.30 M. Add 1 mL of the HCl solution to the Pb(NO3)2 solution. Stir, and wait about 15 seconds. What happens? Record your result.

Step 3 To the Pb(NO3)2 solution add the HCl in 1-mL increments until a noticeable amount of white

solid PbCl2 is present after stirring. Record the total volume of HCl added at that point. Step 4 Put the test tube with the precipitate of PbCl2 into the hot-water bath. Stir for a few moments.

What happens? Record your observations. Cool the test tube under the cold-water tap. What happens? Step 5 Rinse out your graduated cylinder and then add about 5.0 mL of distilled water to the cylin-

der. Add water in 1-mL increments to the mixture in the test tube, stirring well after each addition. Record the volume of water added when the precipitate just dissolves. Answer the questions and do the calculations in Part B before proceeding.

C. Complex Ion Equilibria Many metallic ions in solution exist not as simple ions but, rather, as complex ions in combination with other ions or molecules, called ligands. For example, the Co2+ ion in solution exists as the pink Co(H 2O)6 2+ complex ion, and Cu2+ as the blue Cu(H 2O)4 2+ complex ion. In both of these ions the ligands are H2O molecules. Complex ions are reasonably stable but may be converted to other complex ions on addition of ligands that form more stable complexes than the original ones. Among the common ligands that may form complex species, OH−, NH3, and Cl− are important. An interesting Co(II) complex is the CoCl 4 2− ion, which is blue. This ion is stable in concentrated Cl− solutions. Depending upon conditions, Co(II) in solution may exist as either Co(H 2O)6 2+ or as CoCl 4 2− . The principles of chemical equilibrium can be used to predict which ion will be present: Co(H 2O)6 2+ (aq) + 4 Cl−(aq)

CoCl 4 2− (aq) + 6 H2O

(7)

Step 1 Put a few small crystals (∼0.1 g) of CoCl2 ⋅ 6 H2O in a regular test tube. Add 2 mL of 12 M

HCl. C A U T I O N : This is a concentrated solution of an acidic gas; avoid contact with it and its fumes. This is a concentrated strong acid with a choking odor. Stir to dissolve the crystals. Record the color of the solution. Step 2 Add 2-mL portions of distilled water, stirring after each dilution, until no further color

change occurs. Record the new color.


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Step 3 Place the test tube into the hot-water bath and note any change in color. Cool the tube

under the water tap and report your observations. Complete the questions in Part C before continuing.

D. Dissolving Insoluble Solids We saw in Part B that we can dissolve more PbCl2 by either heating its saturated solution or by simply adding water. In most cases these procedures won’t work very well on other solids because they are typically much less soluble than PbCl2. There are, however, some very powerful methods for dissolving solids that depend on their effectiveness upon the principles of equilibrium. As an example of an insoluble substance, we might consider Zn(OH)2: Zn(OH)2(s)

Zn2+(aq) + 2 OH−(aq)

Ksp = 5 × 10−17 = [Zn2+] × [OH−]−2

(8)

The equilibrium constant for Reaction 8 is very small, which tells us that the reaction does not go very far to the right or, equivalently, that Zn(OH)2 is almost completely insoluble in water. Adding a few drops of a solution containing OH− ion to one containing Zn2+ ion will cause precipitation of Zn(OH)2. At first sight you might well wonder how one could possibly dissolve, say, 1 mole of Zn(OH)2 in an aqueous solution. If, however, you examine Equation 8, you can see, from the equation for Ksp, that in the saturated solution [Zn2+] × [OH−]2 must equal 5 × 10−17. If, by some means, we can lower that product to a value below 5 × 10−17, then Zn(OH)2 will dissolve, until the product becomes equal to Ksp, where equilibrium will again exist. To lower the product, we need to lower the concentration of either Zn2+ or OH− very drastically. This turns out to be very easy to do. To lower [OH−] we can add H+ ions from an acid. If we do that, we drive Reaction 3 to the left, making [OH−] very, very small—small enough to dissolve substantial amounts of Zn(OH)2. To lower [Zn2+] we can take advantage of the fact that zinc(II) forms stable complex ions with both OH− and NH3: Zn2+(aq) + 4 OH−(aq)

Zn(OH)4 2− (aq)

K1 = 3 × 1015

(9)

Zn2+(aq) + 4 NH3(aq)

Zn( NH3 )4 2+ (aq) K2 = 1 × 109

(10)

In high concentrations of OH− ion, Reaction 9 is driven strongly to the right, making [Zn2+] very low. The same thing would happen in solutions containing high concentrations of NH3. In both media we would therefore expect that Zn(OH)2 might dissolve, since if [Zn2+] is very low, Reaction 8 must go to the right. Step 1 To each of three small test tubes add about 2 mL 0.1 M Zn(NO3)2. In this solution [Zn2+]

equals 0.1 M. To each test tube add one drop 6 M NaOH and stir. Report your observations. Step 2 To the first tube add 6 M HCl drop by drop, with stirring. To the second add 6 M NaOH,

again drop by drop. To the third add 6 M NH3. Note what happens in each case. Step 3 Repeat Steps 1 and 2, this time using a solution of 0.1 M Mg(NO3)2. Record your observa-

tions. Answer the questions in Part D. DISPOSAL OF REACTION PRODUCTS. The residues from Parts B and C should be poured in the waste crock. Those from Parts A and D may be poured down the sink.



Name ____________________________________

Section _________________________________

Experiment 22

Data and Observations: Properties of Systems in Chemical Equilibrium

A. Acid-Base Indicators 1. Color of methyl violet in water ____________

2. Reagent causing color change ____________ 3. Reagent causing shift back ____________. Explain, by considering how changes in [H+] will cause Reaction 4 to shift to right and left, why the reagents in Steps 2 and 3 caused the solution to change color. Note that Reactions 3 and 4 must both go to equilibrium after a reagent is added.

B. Solubility Equilibrium; Finding a Value for Ksp 2. Vol. 0.30 M Pb(NO3)2 = 5.0 mL No. moles Pb2+ = [M × V(stock)] = 1.5 × 10−3 moles

Observations: 3. Vol. 0.30 M HCl used ____________ mL No. moles Cl− added ____________ moles

4. Observations: in hot water ____________ in cold water ____________

5. Volume H2O added to dissolve PbCl2 ____________ mL

Total volume of solution ____________ mL

a. Explain why PbCl2 did not precipitate immediately on addition of HCl. (What condition must be met by [Pb2+] and [Cl−] if PbCl2 is to form?)

b. Explain your observations in Step 4. (In which direction did Reaction 5 shift when heated? What must have happened to the value of Ksp in the hot solution? What does this tell you about the sign of ΔH in Reaction 5?)

c. Explain why the PbCl2 dissolved when water was added in Step 5. (What was the effect of the added water on [Pb2+] and [Cl−]? In what direction would such a change drive Reaction 5?)



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Experiment 22


d. Given the numbers of moles of Pb2+ and Cl− in the final solution in Step 5, and the volume of that solution, calculate [Pb2+] and [Cl−] in that solution. [Pb2+] ____________ M; [Cl−] ____________ M

Noting that the molarities just calculated are essentially those in equilibrium with solid PbCl2, calculate [Pb2+] × [Cl−]2. This is equal to Ksp for PbCl2. Ksp = ____________

C. Complex Ion Equilibria 1. Color of CoCl2 ⋅ 6 H2O Color in solution in 12 M HCl

____________

____________

2. Color in diluted solution

____________

3. Color of hot solution

____________

Color of cooled solution

____________

Formula of Co(II) complex ion present in solution in

a. 12 M HCl

____________

b. diluted solution

____________

c. hot solution

____________

d. cooled solution

____________

Explain the color change that occurred when a. water was added in Step 2. (Consider how a change in [Cl−] and [H2O] will shift Reaction 7.) b. the diluted solution was heated. (How would Reaction 7 shift if Kc went up? How did increasing the temperature affect the value of Kc? What is the sign of ΔH in Reaction 7?) (continued on following page) Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Experiment 22


177

D. Dissolving Insoluble Solids 1. Observations on addition of a drop of 6 M NaOH to Zn(NO3)2 solution: 2. Effect on solubility of Zn(OH)2: a. of added HCl solution b. of added NaOH solution c. of added NH3 solution 3. Observations on addition of one drop of 6 M NaOH to Mg(NO3)2 solution: Effect on solubility of Mg(OH)2: a. of added HCl solution b. of added NaOH solution c. of added NH3 solution Explain your observations in Step 1. (Consider Reaction 8; how is it affected by addition of OH− ion?)

In Step 2(a), how does an increase in [H+] affect Reaction 3? (What does that do to Reaction 8?) Explain your observations in Step 2(a).

In Step 2(b), how does an increase in [OH−] affect Reaction 9? What does that do to Reaction 8? Explain your observations in Step 2(b).

In Step 2(c), how does an increase in [NH3] affect Reaction 10? What does that do to Reaction 8? Explain your observations in Step 2(c).

In Step 3, you probably found that Mg(OH)2 was similar in some ways in its behavior to that of Zn(OH)2, but different in others. a. How was it similar? Explain that similarity. (In particular, why would any insoluble hydroxide tend to dissolve in acidic solution?)

b. How was it different? Explain that difference. (In particular, does Mg2+ appear to form complex ions with OH− and NH3? What would we observe if it did? If it did not?)



Name ____________________________________

Section _________________________________

Experiment 22

Advance Study Assignment: Properties of Systems in Chemical Equilibrium 1. Methyl orange, HMO, is a common acid-base indicator. In solution it ionizes according to the equation:

HMO(aq) red

H+(aq) + MO−(aq) yellow

If methyl orange is added to distilled water, the solution turns yellow. If a drop or two of 6 M HCl is added to the yellow solution, it turns red. If to that solution one adds a few drops of 6 M NaOH the color reverts to yellow. a. Why does adding 6 M HCl to the yellow solution of methyl orange tend to cause the color to change to red? (Note that in solution HCl exists as H+ and Cl− ions.)

b. Why does adding 6 M NaOH to the red solution tend to make it turn back to yellow? (Note that in solution NaOH exists as Na+ and OH− ions. How does increasing [OH−] shift Reaction 3 in the discussion section? How would the resulting change in [H+] affect the dissociation reaction of HMO?)

2. Magnesium hydroxide is only very slightly soluble in water. The reaction by which it goes into solution is: Mg(OH)2(s)

Mg2+(aq) + 2 OH−(aq)

a. Formulate the expression for the equilibrium constant, Ksp, for the above reaction.

b. It is possible to dissolve significant amounts of Mg(OH)2 in solutions in which the concentration of either Mg2+ or OH− is very, very small. Explain, using Ksp, why this is the case.

c. Explain why Mg(OH)2 might have very appreciable solubility in 1 M HCl. (Consider the effect of Reaction 3 on the Mg(OH)2 solution reaction.)



Experiment 23 Determination of the Equilibrium Constant for a Chemical Reaction

hen chemical substances react, the reaction typically does not go to completion. Rather, the system goes to some intermediate state in which both the reactants and products have concentrations that do not change with time. Such a system is said to be in chemical equilibrium. When in equilibrium at a particular temperature, a reaction mixture obeys the Law of Chemical Equilibrium, which imposes a condition on the concentrations of reactants and products. This condition is expressed in the equilibrium constant Kc for the reaction. In this experiment we will study the equilibrium properties of the reaction between iron(III) ion and thiocyanate ion:

W

Fe3+(aq) + SCN−(aq)

FeSCN2+(aq)

(1)

When solutions containing Fe3+ ion and thiocyanate ion are mixed, Reaction 1 occurs to some extent, forming the FeSCN2+ complex ion, which has a deep red color. As a result of the reaction, the equilibrium amounts of Fe3+ and SCN− will be less than they would have been if no reaction occurred; for every mole of FeSCN2+ that is formed, one mole of Fe3+ and one mole of SCN− will react. According to the Law of Chemical Equilibrium, the equilibrium constant expression Kc for Reaction 1 is formulated as follows: [FeSCN 2+ ] = Kc [Fe3+ ][SCN − ]

(2)

The value of Kc in Equation 2 is constant at a given temperature. This means that mixtures containing Fe3+ and SCN− will react until Equation 2 is satisfied, so that the same value of the Kc will be obtained no matter what initial amounts of Fe3+ and SCN− were used. Our purpose in this experiment will be to find Kc for this reaction for several mixtures made up in different ways, and to show that Kc indeed has the same value in each of the mixtures. The reaction is a particularly good one to study because Kc is of a convenient magnitude and the color of the FeSCN2+ ion makes for an easy analysis of the equilibrium mixture. The mixtures will be prepared by mixing solutions containing known concentrations of iron(III) nitrate, Fe(NO3)3, and potassium thiocyanate, KSCN. The color of the FeSCN2+ ion formed will allow us to determine its equilibrium concentration. Knowing the initial composition of a mixture and the equilibrium concentration of FeSCN2+, we can calculate the equilibrium concentrations of the rest of the pertinent species and then determine Kc. Since the calculations required in this experiment may not be apparent, we will go through a step-by-step procedure by which they can be made. As a specific example, let us assume that we prepare a mixture by mixing 10.0 mL of 2.00 × 10−3 M Fe(NO3)3 with 10.0 mL of 2.00 × 10−3 M KSCN. As a result of Reaction 1, some red FeSCN2+ ion is formed. By the method of analysis described later, its concentration at equilibrium is found to be 1.50 × 10−4 M. Our problem is to find Kc for the reaction from this information. To do this we first need to find the initial number of moles of each reactant in the mixture. Second, we determine the number of moles of product that were formed at equilibrium. Since the product was formed at the expense of reactants, we can calculate the amount of each reactant that was used up. In the third step we find the number of moles of each reactant remaining in the equilibrium mixture. Fourth, we determine the concentration of each reactant. Finally, in the fifth step we evaluate Kc for the reaction.


182

Experiment 23

Determination of the Equilibrium Constant for a Chemical Reaction

Step 1 Finding the Initial Number of Moles of Each Reactant. This requires relating the vol-

umes and concentrations of the reagent solutions that were mixed to the numbers of moles of each reactant species in those solutions. By the definition of the molarity, MA, of a species A, MA =

no. moles A no. liters of solution, V

or no. moles A = M A × V

(3)

Using Equation 3, it is easy to find the initial number of moles of Fe3+ and SCN−. For each solution the volume used was 10.0 mL, or 0.0100 L. The molarity of each of the solutions was 2.00 × 10−3 M, so MFe3+ = 2.00 × 10−3 M and MSCN − = 2.00 × 10−3 M. Therefore, in the reagent solutions, we find that initial no. moles Fe3+ = MFe3+ × V = 2.00 × 10−3 M × 0.0100 L = 20.0 × 10−6 moles initial no. moles SCN− = MSCN − × V = 2.00 × 10−3 M × 0.0100 L = 20.0 × 10−6 moles Step 2 Finding the Number of Moles of Product Formed. Here again we can use Equation 3 to

advantage. The concentration of FeSCN2+ was found to be 1.50 × 10−4 M at equilibrium. The volume of the mixture at equilibrium is the sum of the two volumes that were mixed, and is 20.0 mL, or 0.0200 L. So, no. moles FeSCN2+ = MFeSCN2+ × V = 1.50 × 10−4 M × 0.0200 L = 3.00 × 10−6 moles The number of moles of Fe3+ and SCN− that were used up in producing the FeSCN2+ must also both be equal to 3.00 × 10−6 moles since, by Equation 1, it takes one mole Fe3+ and one mole SCN− to make each mole of FeSCN2+. Step 3 Finding the Number of Moles of Each Reactant Present at Equilibrium. In Step 1 we

determined that initially we had 20.0 × 10−6 moles Fe3+ and 20.0 × 10−6 moles SCN− present. In Step 2 we found that in the reaction 3.00 × 10−6 moles Fe3+ and 3.00 × 10−6 moles SCN− were used up. The number of moles present at equilibrium must equal the number we started with minus the number that reacted. Therefore, at equilibrium, no. moles at equilibrium = initial no. moles − no. moles used up equil. no. moles Fe3+ = 20.0 × 10−6 − 3.00 × 10−6 = 17.0 × 10−6 moles

(4)

equil. no. moles SCN− = 20.0 × 10−6 − 3.00 × 10−6 = 17.0 × 10−6 moles Step 4 Find the Concentrations of All Species at Equilibrium. Experimentally, we obtained the

concentration of FeSCN2+ directly. [FeSCN2+] = 1.50 × 10−4 M. The concentrations of Fe3+ and SCN− follow from Equation 3. The number of moles of each of these species at equilibrium was obtained in Step 3. The volume of the mixture being studied was 20.0 mL, or 0.0200 L. So, at equilibrium, [Fe3+ ] = MFe3+ =

17.0 × 10 −6 moles no. moles Fe3+ = = 8.50 × 10 −4 M 0.0200 L volume of solution

[SCN − ] = MSCN − =

no. moles SCN − 17.0 × 10 −6 moles = = 8.50 × 10 −4 M volume of solution 0.0200 L

Step 5 Finding the Value of Kc for the Reaction. Once the equilibrium concentrations of all the

reactants and products are known, one needs merely to substitute into Equation 2 to determine Kc: Kc =

[FeSCN 2+ ] 1.50 × 10 −4 = 208 − = 3+ [Fe ][SCN ] (8.50 × 10 −4 ) × (8.50 × 10 −4 )


Experiment 23


183

Step 5 Continued. In this experiment you will obtain data similar to that shown in this example.

The calculations involved in processing that data are completely analogous to those we have made. (Actually, your results will differ from the ones we obtained, since the data in our example were obtained at a different temperature and so relate to a different value of Kc.) In carrying out this analysis we made the assumption that the reaction which occurred was given by Equation 1. There is no inherent reason why the reaction might not have been Fe3+(aq) + 2 SCN−(aq)

Fe(SCN) 2+(aq)

(5)

If you are interested in matters of this sort, you might ask how we know whether we are actually observing Reaction 1 or Reaction 5. The line of reasoning is the following. If Reaction 1 is occurring, Kc for that reaction as we calculate it should remain constant with different reagent mixtures. If, however, Reaction 5 is going on, Kc as calculated for that reaction should remain constant. In the optional part of the Data and Calculations section, we will assume that Reaction 5 occurs and make the analysis of Kc on that basis. The results of the two sets of calculations should make it clear that Reaction 1 is the one that we are studying. Two analytical methods can be used to determine [FeSCN2+] in the equilibrium mixtures. The more precise method uses a spectrophotometer, which measures the amount of light absorbed by the red complex at 447 nm, the wavelength at which the complex most strongly absorbs. The absorbance, A, of the complex is proportional to its concentration, M, and can be measured directly on the spectrophotometer: A = kM

(6)

Your instructor will show you how to operate the spectrophotometer, if one is available in your laboratory, and will provide you with a calibration curve or equation from which you can find [FeSCN2+] once you have determined the absorbance of your solutions. See Appendix IV for information about spectrophotometers. In the other analytical method a solution of known concentration of FeSCN2+ is prepared. The [FeSCN2+] concentrations in the solutions being studied are found by comparing the color intensities of these solutions with that of the known. The method involves matching the color intensity of a given depth of unknown solution with that for an adjusted depth of known solution. The actual procedure and method of calculation are discussed in the Experimental Procedure section. In preparing the mixtures in this experiment we will maintain the concentration of H+ ion at 0.5 M. The hydrogen ion does not participate directly in the reaction, but its presence is necessary to avoid the formation of brown-colored species such as FeOH2+, which would interfere with the analysis of [FeSCN2+].

Experimental Procedure Label five regular test tubes (18 mm × 150 mm) 1 to 5, with labels or by noting their positions on your test tube rack. Pour about 30 mL 2.00 × 10−3 M Fe(NO3)3 in 1 M HNO3 into a dry 100-mL beaker. Pipet 5.00 mL of that solution into each test tube. Then add about 20 mL 2.00 × 10−3 M KSCN to another dry 100-mL beaker. Pipet 1, 2, 3, 4, and 5 mL from the KSCN beaker into each of the corresponding test tubes labeled 1 to 5. Then pipet the proper number of milliliters of water into each test tube to bring the total volume in each tube to 10.00 mL. The volumes of reagents to be added to each tube are summarized in Table 23.1, which you should complete by filling in the required volumes of water. See Appendix IV for a discussion of the use of pipets. Mix each solution thoroughly with a glass stirring rod. Be sure to dry the stirring rod after mixing each solution.

Method I. Analysis by Spectrophotometric Measurement Place a portion of the mixture in tube 1 in a spectrophotometer cell, as demonstrated by your instructor, and measure the absorbance of the solution at 447 nm. Determine the concentration of FeSCN2+ from the calibration curve provided for each instrument or from the equation furnished to you. Record the value on the Data


184

Experiment 23


Table 23.1 Test Tube No.

Volume Fe(NO3)3 solution (mL) Volume KSCN solution (mL) Volume H2O (mL)

1

2

3

4

5

5.00 1.00 —

5.00 2.00 —

5.00 3.00 —

5.00 4.00 —

5.00 5.00 —

page. Repeat the measurement using the mixtures in each of the other test tubes. For a discussion of how absorbance and concentration are related, see Appendix IV.

Method II. Analysis by Comparison with a Standard Prepare a solution of known [FeSCN2+] by pipetting 10.00 mL 0.200 M Fe(NO3)3 in 1 M HNO3 into a test tube and adding 2.00 mL 0.00200 M KSCN and 8.00 mL water. Mix the solution thoroughly with a stirring rod. Since in this solution [Fe3+] >> [SCN−], Reaction 1 is driven strongly to the right. You can assume without serious error that essentially all the SCN− added is converted to FeSCN2+. Assuming that this is the case, calculate [FeSCN2+] in the standard solution and record the value on the Data page. The [FeSCN2+] in the unknown mixture in test tubes 1 to 5 can be found by comparing the intensity of the red color in these mixtures with that of the standard solution. This can be done by placing the test tube containing Mixture 1 next to a test tube containing the standard. Look down both test tubes toward a wellilluminated piece of white paper on the laboratory bench. Pour out the standard solution into a dry, clean beaker until the color intensity you see down the tube containing the standard matches that which you see when looking down the tube containing the unknown. Use a well-lit piece of white paper as your background. When the colors match, the following relation is valid: [FeSCN2+]unknown × depth of unknown solution = [FeSCN2+] × depth of standard solution

(7)

Measure the depths of the matching solutions with a rule and record them. Repeat the measurement for Mixtures 2 through 5, recording the depth of each unknown and that of the standard solution which matches it in intensity. DISPOSAL OF REACTION PRODUCTS. In this experiment, reactant concentrations are very low. In most localities you can pour the contents of the test tubes down the sink when you have completed your measurements. However, consult your instructor for alternate disposal procedures.


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Section ________________________________

5.00

5.00

5.00

5.00

2

3

4

5

5.00

4.00

3.00

2.00

1.00

Volume in mL, 2.00 × 10−3 M KSCN

________

________

________

________

________

Volume in mL, Water

________

________

________

________

________

Method I Absorbance

________

________

________

________

________

Standard

Fe3+(aq) + SCN−(aq)

FeSCN2+(aq)

________ × 10−4 M

________ × 10−4 M

________ × 10−4 M

________ × 10−4 M

________ × 10−4 M

[FeSCN]2+

(1)


to find the number of moles of FeSCN2+ in each of the mixtures, and enter the values in the fifth column of the table. Note that this is also the number of moles of Fe3+ and SCN− that were used up in the reaction.

Step 2 Enter the experimentally determined value of [FeSCN2+] at equilibrium for each of the mixtures in the next to last column in the table. Use Equation 3

the table.

Step 1 Find the initial number of moles of Fe3+ and SCN− in the mixtures in test tubes 1 through 5. Use Equation 3 and enter the values in the first two columns of

This calculation is most easily done by following Steps 1 through 5 in the discussion. Results are to be entered in the table on the following page. If you are using Excel, set up the table as we have, and follow the directions that follow.

A. Calculation of Kc assuming the reaction:

Processing the Data

________

________

________

________

________

Unknown

Method II Depth in mm

If Method II was used, [FeSCN2+]standard = ______________ × 10−4 M; [FeSCN2+] in Mixtures 1 to 5 is found by Equation 7.

5.00

1

Mixture

Volume in mL, 2.00 × 10−3 M Fe(NO3)3

Data and Calculations: Determination of the Equilibrium Constant for a Chemical Reaction

Experiment 23

Name ____________________________________

_______

_______

_______

_______

2

3

4

5

_______

_______

_______

_______

_______ × 10−6

SCN-

Mean value of Kc _____________

_______ × 10−6

Fe3+

Initial No. Moles

1

Mixture

_______

_______

_______

_______

_______ × 10−6

SCN-

_______

_______

_______

_______

_______ × 10−6

FeSCN2+

Standard deviation ______________

_______

_______

_______

_______

_______ × 10−6

Fe3+

Equilibrium No. Moles

_______

_______

_______

_______

_______ × 10−4 M

[Fe3+]

_______

_______

_______

_______

_______ × 10−4 M

[SCN-]

Equilibrium Concentrations

_______

_______

_______

_______

_______


_______

_______

_______

_______

_______ × 10−4 M

[FeSCN2+]

Kc

Experiment 23

Step 6 Calculate the mean value for Kc and the standard deviation. (See Appendix VIII.)

Step 5 Calculate Kc for the reaction for each of the mixtures by substituting values for the equilibrium concentrations of Fe3+, SCN−, and FeSCN2+ in Equation 2.

0.0100 liter in all cases. Enter the values in columns 6 and 7 of the table.

Step 4 Use Equation 3 and the results of Step 3 to find the concentrations of all of the species at equilibrium. The volume of the mixture is 10.00 mL, or

culate the number of moles of Fe3+ and SCN− that remain in each mixture at equilibrium. Use Equation 4. Enter the results in columns 3 and 4 of the table.

Step 3 From the number of moles of Fe3+ and SCN− initially present in each mixture, and the number of moles of Fe3+ and SCN− used up in forming FeSCN2+, cal-

186 Determination of the Equilibrium Constant for a Chemical Reaction



Fe(SCN) 2+(aq)

(5)

_______

_______

3

5

_______

_______

_______

SCN-

_______

_______

_______

Fe3+

_______

_______

_______

SCN-

Equilibrium No. Moles

_______

_______

_______

Fe(SCN)2+

_______

_______

_______

[Fe3+]

_______

_______

_______

[SCN-]

Equilibrium Concentrations

_______

_______

_______

[Fe(SCN)2+]

_______

_______

_______

Kc

On the basis of the results of Part A, what can you conclude about the validity of the equilibrium concept, as exemplified by Equation 2? What do you conclude about the formula of iron(III) thiocyanate complex ion?

_______

Fe3+

Initial No. Moles

1

Mixture

If we analyze the equilibrium system we have studied, assuming that Reaction 5 occurs rather than Reaction 1, we would presumably obtain nonconstant values of Kc. Using the same kind of procedure as in Part A, calculate Kc for Mixtures 1, 3, and 5 on the basis that Fe(SCN) 2+ is the formula of the complex ion formed by the reaction between Fe3+ and SCN−. As a result of the procedure used for calibrating the system by Method I or Method II, [Fe(SCN) 2+] will equal one-half the [FeSCN2+] obtained for each solution in Part A. Note that two moles SCN2 are needed to form one mole Fe(SCN)2 +. This changes the expression for Kc. Also, in calculating the equilibrium number of moles SCN− you will need to subtract (2 × number of moles Fe(SCN)2 +) from the initial number of moles SCN−.

Fe3+(aq) + 2 SCN−(aq)

B. In calculating Kc in Part A, we assume, correctly, that the formula of the complex ion is FeSCN2+. It is by no means obvious that this is the case and one might have assumed, for instance, that Fe(SCN) 2+ was the species formed. The reaction would then be

Experiment 23 Determination of the Equilibrium Constant for a Chemical Reaction

187




Name ____________________________________

Section ________________________________

Experiment 23

Advance Study Assignment: Determination of the Equilibrium Constant for a Chemical Reaction 1. A student mixes 5.00 mL 2.00 × 10−3 M Fe(NO3)3 with 3.00 mL 2.00 × 10−3 M KSCN. She finds that in the equilibrium mixture the concentration of FeSCN2+ is 1.28 × 10−4 M. Find Kc for the reaction Fe3+(aq) + SCN−(aq) FeSCN2+(aq). Step 1 Find the number of moles Fe3+ and SCN− initially present. (Use Eq. 3.)

____________ moles Fe3+; ____________ moles SCN− Step 2 How many moles of FeSCN2+ are in the mixture at equilibrium? What is the volume of the

equilibrium mixture? (Use Eq. 3.)

____________ mL; ____________ moles FeSCN2+ How many moles of Fe3+ and SCN− are used up in making the FeSCN2+?

____________ moles Fe3+; ____________ moles SCN− Step 3 How many moles of Fe3+ and SCN− remain in the solution at equilibrium? (Use Eq. 4 and

the results of Steps 1 and 2.) ____________ moles Fe3+; ____________ moles SCN− Step 4 What are the concentrations of Fe3+, SCN−, and FeSCN2+ at equilibrium? What is the volume

of the equilibrium mixture? (Use Eq. 3 and the results of Step 3.)

[Fe3+] = ____________ M; [SCN−] = ____________ M; [FeSCN2+] = ____________ M

____________ mL Step 5 What is the value of Kc for the reaction? (Use Eq. 2 and the results of Step 4.)

Kc = ____________ (continued on following page)


190 2.

Experiment 23


Assume that the reaction studied in Problem 1 is Fe3+(aq) + 2 SCN−(aq) Fe(SCN)2+(aq). Find Kc for this reaction, given the data in Problem 1, except that the equilibrium concentration of Fe(SCN)2+ is equal to 0.49 × 10−4 M. Optional

a. Formulate the expression for Kc for the alternate reaction just cited. b. Find Kc as you did in Problem 1; take due account of the fact that two moles SCN− are used up per mole Fe(SCN)2 + formed. Step 1 Results are as in Problem 1. Step 2 How many moles of Fe(SCN)2 + are in the mixture at equilibrium? (You should use Eq. 3.)

____________ moles Fe(SCN)2+ How many moles of Fe3+ and SCN− are used up in making the Fe(SCN)2 +?

____________ moles Fe3+; ____________ moles SCN− Step 3 How many moles of Fe3+ and SCN− remain in solution at equilibrium? Use the results of

Steps 1 and 2, noting that no. moles SCN− at equilibrium = original no. moles SCN− − (2 × no. moles Fe(SCN)2 +).

____________ moles Fe3+; ____________moles SCN− Step 4 What are the concentrations of Fe3+, SCN−, and Fe(SCN)2+ at equilibrium? (Use Eq. 3 and

the results of Step 3.)

[Fe3+] = __________ M; [SCN−] = __________ M; [Fe(SCN)2+] = __________ M Step 5 Calculate Kc on the basis that the alternate reaction occurs. (Use the answer to Part 2a.)

Kc = ____________


Experiment 24 The Standardization of a Basic Solution and the Determination of the Molar Mass of an Acid

W

hen a solution of a strong acid is mixed with a solution of a strong base, a chemical reaction occurs that can be represented by the following net ionic equation: H+(aq) + OH−(aq) → H2O

This is called a neutralization reaction, and chemists use it extensively to change the acidic or basic properties of solutions. The equilibrium constant for the reaction is about 1014 at room temperature, so that the reaction can be considered to proceed completely to the right, using up whichever of the ions is present in the lesser amount and leaving the solution either acidic or basic, depending on whether H+ or OH− ion was in excess. Since the reaction is essentially quantitative, it can be used to determine the concentrations of acidic or basic solutions. A frequently used procedure involves the titration of an acid with a base. In the titration, a basic solution is added from a buret to a measured volume of acid solution until the number of moles of OH− ion added is just equal to the number of moles of H+ ion present in the acid. At that point the volume of basic solution that has been added is measured. Recalling the definition of the molarity, MA, of species A, we have MA =

no. moles of A no. liters of solution, V

or no. moles of A = MA × V

(1)

At the end point of the titration, no. moles H+ originally present = no. moles OH− added

(2)

MH + × Vacid = MOH − × Vbase

(3)

So, by Equation 1,

Therefore, if the molarity of either the H+ or the OH− ion in its solution is known, the molarity of the other ion can be found from the titration. The equivalence point or end point in the titration is determined by using a chemical, called an indicator, that changes color at the proper point. The indicators used in acid-base titrations are weak organic acids or bases that change color when they are neutralized. One of the most common indicators is phenolphthalein, which is colorless in acid solutions but becomes red when the pH of the solution becomes 9 or higher. When a solution of a strong acid is titrated with a solution of a strong base, the pH at the end point will be about 7. At the end point a drop of acid or base added to the solution will change its pH by several pH units, so that phenolphthalein can be used as an indicator in such titrations. If a weak acid is titrated with a strong base, the pH at the equivalence point is somewhat higher than 7, perhaps 8 or 9, and phenolphthalein is still a very satisfactory indicator. If, however, a solution of a weak base such as ammonia is titrated with a strong acid, the pH will be a unit or two less than 7 at the end point, and phenolphthalein will not be as good an indicator for that titration as, for example, methyl red, whose color changes from red to yellow as the pH changes from about 4 to 6. Ordinarily, indicators will be chosen so that their color change occurs at about the pH at the equivalence point of a given acid-base titration.


192

Experiment 24

The Standardization of a Basic Solution and the Determination of the Molar Mass of an Acid

In this experiment you will determine the molarity of OH− ion in an NaOH solution by titrating that solution against a standardized solution of HCl. Since in these solutions one mole of acid in solution furnishes one mole of H+ ion and one mole of base produces one mole of OH− ion, MHCl = MH + in the acid solution, and MNaOH = MOH − in the basic solution. Therefore, the titration will allow you to find MNaOH as well as MOH − . In the second part of this experiment you will use your standardized NaOH solution to titrate a sample of a pure solid organic or inorganic acid. By titrating a weighed sample of the unknown acid with your standardized NaOH solution you can, by Equation 2, find the number of moles H+ ion that your sample can furnish. If your acid has one acidic hydrogen atom in the molecule, with formula HB, then the number of moles of acid will equal the number of moles of H+ that react during the titration. The molar mass of the acid, MM, will equal the number of grams of acid that contain one mole of H+ ion. MM =

no. grams of acid no. moles of H + ion furnished

(4)

Many acids release one mole of H+ ion per mole of acid on titration with NaOH solution. Such acids are called monoprotic. Acetic acid, HC2H3O2, is a classic example of a monoprotic acid (only the first H atom in the formula is acidic). Like all organic acids, acetic acid is weak, in that it only ionizes to a small extent in water solution. Some acids contain more than one acidic hydrogen atom in the molecule, and would have the general formula H2B or H3B. Sulfurous acid, H2SO3, is an example of an inorganic diprotic acid. Maleic acid, H2C4H2O4, is a diprotic organic acid. If we should titrate samples of these acids with a solution of a strong base like NaOH, it would take two moles of OH− ion, or two moles of NaOH, to neutralize one mole of acid, since each mole of acid would release two moles of H+ ion. If you don’t know the formula of an acid, you can’t be sure it is monoprotic, so you can only calculate the mass of acid that will react with one mole of OH− ion. That mass will contain one mole of H+ ion and is called the equivalent mass of the acid. The molar mass and the equivalent mass are related by simple equations. Since sulfurous acid gives up two moles of H+ ion in titration with NaOH, the molar mass must equal twice the equivalent mass, which gives up one mole of H+ ion. To simplify matters, in this experiment we will only use monoprotic acids, with formula HB, so one mole of acid will react with one mole of NaOH, and you can find the molar mass of your acid by Equation 4.

Experimental Procedure Note: This experiment is relatively long unless you know precisely what to do. Study the experiment carefully before coming to class, so that you don’t have to spend a lot of time finding out what the experiment is all about. Obtain two burets and a sample of solid unknown acid from the stockroom.

A. Standardization of NaOH Solution Into a small graduated cylinder draw about 7 mL of the stock 6 M NaOH solution provided in the laboratory and dilute to about 400 mL with distilled water in a 500-mL Florence flask. Stopper the flask tightly and mix the solution thoroughly at intervals over a period of at least 15 minutes before using the solution. Draw into a clean, dry 125-mL Erlenmeyer flask about 75 mL of standardized HCl solution (about 0.1 M) from the stock solution on the reagent shelf. This amount should provide all the standard acid you will need; do not waste it. Record the molarity of the HCl. Prepare for the titration by using the procedure described in Experiment 7 and Appendix IV. The purpose of this procedure is to make sure that the solution in each buret has the same molarity as it has in the container from which it was poured. Clean the two burets and rinse them with distilled water. Then rinse one buret three times with a few milliliters of the HCl solution, in each case thoroughly wetting the walls of the buret


Experiment 24


193

with the solution and then letting it out through the stopcock. Fill the buret with HCl; open the stopcock momentarily to fill the tip. Proceed to clean and fill the other buret with your NaOH solution in a similar manner. Put the acid buret, A, on the left side of your buret clamp, and the base buret, B, on the right side. Check to see that your burets do not leak and that there are no air bubbles in either buret tip. Read and record the levels in the two burets to 0.02 mL. Draw about 25 mL of the HCl solution from the buret into a clean 250-mL Erlenmeyer flask; add to the flask about 25 mL of distilled H2O and two or three drops of phenolphthalein indicator solution. Place a white sheet of paper under the flask to aid in the detection of any color change. Add the NaOH solution intermittently from its buret to the solution in the flask, noting the pink phenolphthalein color that appears and disappears as the drops hit the liquid and are mixed with it. Swirl the liquid in the flask gently and continuously as you add the NaOH solution. When the pink color begins to persist, slow down the rate of addition of NaOH. In the final stages of the titration add the NaOH drop by drop until the entire solution just turns a pale pink color that will persist for about 30 seconds. If you go past the end point and obtain a red solution, add a few drops of the HCl solution to remove the color, and then add NaOH a drop at a time until the pink color persists. Carefully record the final readings on the HCl and NaOH burets. To the 250-mL Erlenmeyer flask containing the titrated solution, add about 10 mL more of the standard HCl solution. Titrate this as before with the NaOH to an end point, and carefully record both buret readings once again. To this solution add about 10 mL more HCl and titrate a third time with NaOH. You have now completed three titrations, with total HCl volumes of about 25, 35, and 45 mL. Using Equation 3, calculate the molarity of your base, MOH − for each of the three titrations. In each case, use the total volumes of acid and base that were added up to that point. At least two of these molarities should agree to within 1%. If they do, proceed to the next part of the experiment. If they do not, repeat these titrations until two calculated molarities do agree. Calculate the mean value of MOH– and the standard deviation. (See Appendix VIII.)

B. Determination of the Molar Mass of an Acid Weigh the vial containing your solid acid on the analytical balance to ±0.0001 g. Carefully pour out about half the sample into a clean but not necessarily dry 250-mL Erlenmeyer flask. Weigh the vial accurately. Add about 50 mL of distilled water and two or three drops of phenolphthalein to the flask. The acid may be relatively insoluble, so don’t worry if it doesn’t all dissolve. Fill your NaOH buret with the (now standardized) NaOH solution. Add the standard HCl to your HCl buret until it is about half full. Read both levels carefully and record them. Titrate the solution of the solid acid with NaOH. As the acid is neutralized it will tend to dissolve in the solution. If the acid appears to be relatively insoluble, add NaOH until the pink color persists, and then swirl to dissolve the solid. If the solid still will not dissolve, and the solution remains pink, add 25 mL of ethanol to increase the solubility. If you go past the end point, add HCl as necessary. The final pink end point should appear on addition of one drop of NaOH. Record the final levels in the NaOH and HCl burets. Pour the rest of your acid sample into a clean 250-mL Erlenmeyer flask, and weigh the vial accurately. Titrate this sample of acid as before with the NaOH and HCl solutions. If you use HCl in these titrations, and you probably will, the calculations needed are a bit more complicated than in the standardization of the NaOH solution. To find the number of moles of H+ ion in the solid acid, you must subtract the number of moles of HCl used from the number of moles of NaOH. For a backtitration, which is what we have in this case: no. moles H+ in solid acid = no. moles OH− in NaOH soln. − no. moles H+ in HCl soln.

(5)

For volumes in milliliters, this equation takes the form: no. moles H+ in solid acid =

MNaOH × VNaOH MHCl × VHCl − 1000 1000

(6)


194

Experiment 24


C. Determination of Ka for the Unknown Acid

Optional

Your instructor will tell you in advance if you are to do this part of the experiment. If you do this part, you will need to save one of the titrated solutions from Part B. Given that titrated solution, it is easy to prepare one in which [H+] ion is equal to Ka for your acid. A weak acid, HB, dissociates in water solution according to the equation: HB(aq)

H+(aq) + B−(aq)

(7)

The equilibrium constant for this reaction is called the acid dissociation constant for HB and is given the symbol Ka. A solution of HB will obey the equilibrium condition given by the equation: Ka =

[H + ][B− ] [HB]

(8)

Your acid is a weak acid, so will follow both of the above equations. In the titration in Part B you converted a solution of HB into one containing NaB by adding NaOH to the end point. If, to your titrated solution, you now add some HCl, it will convert some of the B− ions in the NaB solution back to HB. If the number of moles of HCl you add equals one-half the number of moles of H+ in your original sample, then half of the B− ions will be converted to HB. In the resulting solution, [HB] will equal [B−], and so, by Equation 8, Ka = [H+] in the half-neutralized solution

(9)

Using the approach we have outlined, use your titrated solution to prepare one in which [HB] equals [B−]. Measure the pH of that solution with a pH meter, using the procedure described in Appendix IV. From that value, calculate Ka for your unknown acid. Optional Experiment: Using the procedure in this experiment, find the mass percent acetic acid in household vinegar. What is the molarity of the acid? DISPOSAL OF REACTION PRODUCTS.

The reaction products in this experiment may be diluted

and poured down the drain.


Name ____________________________________

Section ________________________________

Experiment 24

Data and Calculations: The Standardization of a Basic Solution and the Determination of the Molar Mass of an Acid

A. Standardization of NaOH Solution Trial 1 Initial reading, HCl buret

____________ mL

Final reading, HCl buret

____________ mL

Initial reading, NaOH buret

____________ mL

Final reading, NaOH buret

____________ mL

Trial 2

Trial 3

____________ mL

____________ mL

____________ mL

____________ mL

B. Determination of the Molar Mass of an Unknown Acid Mass of vial plus contents

____________ g

Mass of vial plus contents less Sample 1

____________ g

Mass less Sample 2

____________ g Trial 1

Trial 2

Initial reading, NaOH buret

____________ mL

____________ mL

Final reading, NaOH buret

____________ mL

____________ mL

Initial reading, HCl buret

____________ mL

____________ mL

Final reading, HCl buret

____________ mL

____________ mL

Trial 1

Trial 2

Trial 3

Total volume HCl

____________ mL

____________ mL

____________ mL

Total volume NaOH

____________ mL

____________ mL

____________ mL

Processing the Data A. Standardization of NaOH Solution



196

Experiment 24


Molarity, MHCl, of standardized HCl

____________ M

Molarity, MH + , in standardized HCl

____________ M

By Equation 3, MH + × Vacid = MOH − × Vbase

or

MOH − = MH + ×

VHCl VNaOH

(3)

Use Equation 3 to find the molarity, MOH − , of the NaOH solution. Note that the volumes do not need to be converted to liters, since we use the volume ratio. Trial 1 MOH −

____________ M

Trial 2 ____________ M

Trial 3 ____________ M

(should agree within 1%)

The molarity of the NaOH will equal MOH − , since one mole NaOH → one mole OH−. Average molarity of NaOH solution, MNaOH

____________ M

Standard deviation ____________ M

B. Determination of the Molar Mass of the Unknown Acid Mass of sample

Trial 1 ____________ g

Trial 2 ____________ g

Volume NaOH used

____________ mL

____________ mL

____________

____________

____________ mL

____________ mL

____________

____________

____________

____________

____________ g

____________ g

No. moles NaOH =

VNaOH × MNaOH 1000

Volume HCl used No. moles HCl =

VHCl × MHCl 1000

No. moles H+ in sample (use Eq. 5) MM =

no. grams acid no. moles H +

Unknown no.

C. Determination of Ka for the Unknown Acid

____________

Optional

No. moles H+ in sample (from Part B)

____________

No. moles HCl to be added ____________ Volume of HCl added ____________ mL pH of half-neutralized solution ____________ Ka of acid ____________


Name ____________________________________

Section ________________________________

Experiment 24

Advance Study Assignment: Molar Mass of an Acid 1. 7.0 mL of 6.0 M NaOH are diluted with water to a volume of 400 mL. You are asked to find the molarity of the resulting solution. a. First find out how many moles of NaOH there are in 7.0 mL of 6.0 M NaOH. Use Equation 1. Note that the volume must be in liters. ____________ moles b. Since the total number of moles of NaOH is not changed on dilution, the molarity after dilution can also be found by Equation 1, using the final volume of the solution. Calculate that molarity. ____________ M 2. In an acid-base titration, 23.91 mL of an NaOH solution are needed to neutralize 24.58 mL of a 0.1002 M HCl solution. To find the molarity of the NaOH solution, we can use the following procedure: a. First note the value of MH + in the HCl solution. ____________ M b. Find MOH − in the NaOH solution. (Use Eq. 3.) ____________ M c. Obtain MNaOH from MOH − . ____________ M 3. A 0.3174 g sample of an unknown acid requires 26.23 mL of 0.1056 M NaOH for neutralization to a phenolphthalein end point. There are 0.27 mL of 0.1096 M HCl used for back-titration. a. How many moles of OH− are used? How many moles of H+ from HCl? ____________ moles OH− ____________ moles H+ b. How many moles of H+ are there in the solid acid? (Use Eq. 5.) ____________ moles H+ in solid c. What is the molar mass of the unknown acid? (Use Eq. 4.) ____________ g/mol



Experiment 25 pH Measurements—Buffers and Their Properties

ne of the more important properties of an aqueous solution is its concentration of hydrogen ion. The H+ or H3O+ ion has great effect on the solubility of many inorganic and organic species, on the nature of complex metallic cations found in solutions, and on the rates of many chemical reactions. It is important that we know how to measure the concentration of hydrogen ion and understand its effect on solution properties. For convenience, the concentration of H+ ion is frequently expressed as the pH of the solution rather than as molarity. The pH of a solution is defined by the following equation:

O

pH = −log[H+]

(1)

where the logarithm is taken to the base 10. If [H+] is 1 × 10−4 moles per liter, the pH of the solution is, by the equation, 4. If the [H+] is 5 × 10−2 M, the pH is 1.3. Basic solutions can also be described in terms of pH. In water solutions the following equilibrium relation will always be obeyed: [H+] × [OH−] = Kw = 1 × 10−14 at 25°C

(2)

In distilled water [H+] equals [OH−], so, by Equation 2, [H+] must be 1 × 10−7 M. Therefore, the pH of distilled water is 7. Solutions in which [H+] > [OH−] are said to be acidic and will have a pH < 7; if [H+] < [OH−], the solution is basic and its pH > 7. A solution with a pH of 10 will have a [H+] of 1 × 10−10 M and a [OH−] of 1 × 10−4 M. We measure the pH of a solution experimentally in two ways. In the first of these we use a chemical called an indicator, which is sensitive to pH. These substances have colors that change over a relatively short pH range (about two pH units) and can, when properly chosen, be used to determine roughly the pH of a solution. Two very common indicators are litmus, usually used on paper, and phenolphthalein, the most common indicator in acid-base titrations. Litmus changes from red to blue as the pH of a solution goes from about 6 to about 8. Phenolphthalein changes from colorless to red as the pH goes from 8 to 10. A given indicator is useful for determining pH only in the region in which it changes color. Indicators are available for measurement of pH in all the important ranges of acidity and basicity. By matching the color of a suitable indicator in a solution of known pH with that in an unknown solution, one can determine the pH of the unknown to within about 0.3 pH units. The other method for finding pH is with a device called a pH meter. In this device two electrodes, one of which is sensitive to [H+], are immersed in a solution. The potential between the two electrodes is related to the pH. The pH meter is designed so that the scale will directly furnish the pH of the solution. A pH meter gives much more precise measurement of pH than does a typical indicator and is ordinarily used when an accurate determination of pH is needed. Some acids and bases undergo substantial ionization in water, and are called strong because of their essentially complete ionization in reasonably dilute solutions. Other acids and bases, because of incomplete ionization (often only about 1% in 0.1 M solution), are called weak. Hydrochloric acid, HCl, and sodium hydroxide, NaOH, are typical examples of a strong acid and a strong base. Acetic acid, HC2H3O2, and ammonia, NH3, are classic examples of a weak acid and a weak base. A weak acid will ionize according to the Law of Chemical Equilibrium: HB(aq) At equilibrium,

H+(aq) + B−(aq) [H + ][B− ] = Ka [HB]

(3)

(4)


200

Experiment 25

pH Measurements—Buffers and Their Properties

Ka is a constant characteristic of the acid HB; in solutions containing HB, the product of concentrations in the equation will remain constant at equilibrium independent of the manner in which the solution was made. A similar relation can be written for solutions of a weak base. The value of the ionization constant Ka for a weak acid can be found experimentally in several ways. In Experiment 24 we used a very simple method for finding Ka. In general, however, we need to find the concentrations of each of the species in Equation 4 by one means or another. In this experiment we will determine Ka for a weak acid in connection with our study of the properties of those solutions we call buffers. Salts that can be formed by the reaction of strong acids and bases—such as NaCl, KBr, or NaNO3— ionize completely but do not react with water when in solution. They form neutral solutions with a pH of about 7. When dissolved in water, salts of weak acids or weak bases furnish ions that tend to react to some extent with water, producing molecules of the weak acid or base and liberating some OH− or H+ ion to the solution. If HB is a weak acid, the B− ion produced when NaB is dissolved in water will react with water to some extent, according to the equation B−(aq) + H2O

HB(aq) + OH−(aq)

(5)

Solutions of sodium acetate, NaC2H3O2, the salt formed by reaction of sodium hydroxide with acetic acid, will be slightly basic because of the reaction of C2 H3O2 − ion with water to produce HC2H3O2 and OH−. Because of the analogous reaction of the NH 4 + ion with water to form H3O+ ion, solutions of ammonium chloride, NH4Cl, will be slightly acidic. Salts of most transition metal ions are acidic. A solution of CuSO4 or FeCl3 will typically have a pH equal to 5 or lower. The salts are completely ionized in solution. The acidity comes from the fact that the cation is hydrated (e.g., Cu(H 2O)4 2+ or Fe(H2O)63+ ). The large + charge on the metal cation attracts electrons from the O---H bonds in water, weakening them and producing some H+ ions in solution; with CuSO4 solutions the reaction would be Cu(H 2O)4 2− (aq)

Cu(H2O)3OH+(aq) + H+(aq)

(6)

Buffers Some solutions, called buffers, are remarkably resistant to pH changes. Water is not a buffer, since its pH is very sensitive to addition of any acidic or basic species. Even bubbling your breath through a straw into distilled water can lower its pH by at least one pH unit, just due to the small amount of CO2, which is acidic, in exhaled air. With a good buffer solution, you could blow exhaled air into it for half an hour and not change the pH appreciably. All living systems contain buffer solutions, since stability of pH is essential for the occurrence of many of the biochemical reactions that go on to maintain the living organism. There is nothing mysterious about what one needs to make a buffer. All that is required is a solution containing a weak acid and its conjugate base. An example of such a solution is one containing the weak acid HB, and its conjugate base, B− ion, obtained by dissolving the salt NaB in water. The pH of such a buffer is established by the relative concentrations of HB and B− in the solution. If we manipulate Equation 4, we can solve for the concentration of H+ ion: [H+] = Ka × [HB]/[B−]

(4a)

This equation is often further modified when working with buffers, by taking the negative logarithm to base 10 of both sides: pH = pKa + log [B−]/[HB]

(4b)

where pKa equals −log Ka. This equation is called the Henderson-Hasselbach equation, familiar to biochemists. Equation 4b tells us what quantities fix the pH of a buffer.


Experiment 25


201

If we are working with a weak acid HB whose Ka equals 1 × 10−5, its pKa equals 5.0. If we mix equal volumes of 0.10 M HB and 0.10 M NaB, the pH equals pKa equals 5.0, since the concentrations of HB and B− are equal in the final solution, and log 1 equals 0. [H+] equals 1 × 10−5 M. The H+ ion in the solution has to come from ionization of HB, but the amount is tiny, so we can assume that in any buffer the weak acid and its conjugate base do not react appreciably with one another when their solutions are mixed. Their relative concentrations can be calculated from the way the buffer was put together. Using Equation 4b, you should be able to answer several questions regarding buffers. For example, how does the pH of buffer solution change if you dilute it with water? How does it change if you add an HB solution? What happens if you add a solution of NaB, containing the conjugate base? What is the pH range over which a buffer would be useful, if you assume that the ratio of [B−] to [HB] must lie between 10:1 and 1:10? How sensitive is the pH of water to the addition of a strong acid, like HCl, or addition of a strong base, like NaOH? What happens when you add a strong acid or a strong base to a buffer? Why does the pH of a buffer resist changing when small amounts of a strong acid or strong base are added? Some of these questions you can answer by just looking at Equation 4b. For others you need to realize that, although a weak acid and its conjugate base don’t react with each other, a weak acid like HB will react quantitatively in a buffer with a strong base, like NaOH: HB (aq) + OH− (aq) → B− (aq) + H2O

(7)

and a weak base like NaB will react with a strong acid like HCl: B− (aq) + H+ (aq) → HB (aq)

(8)

In Reaction 7, a small amount of NaOH will not raise the pH very much, since the OH− ion is essentially soaked up by the acid HB, producing some B− ion and increasing the value of [B−]/[HB], but not destroying the buffer. Reactions 7 and 8 can also be used to prepare a buffer from a solution of HB by addition of NaOH, or from a solution of NaB by addition of HCl. In this experiment you will determine the approximate pH of several solutions by using acid-base indicators. Then you will find the pH of some other solutions with a pH meter. In the rest of the experiment you will carry out some reactions that will allow you to answer all the questions we have raised about buffers. Finally, you will prepare one or two buffers having specific pH values.

Experimental Procedure You may work in pairs on the first three parts of this experiment.

A. Determination of pH by the Use of Acid-Base Indicators To each of five small test tubes add about 1 mL 0.10 M HCl (about 1/2-inch depth in tube). To each tube add a drop or two of one of the indicators mentioned in Table 25.1, one indicator to a tube. Note the color of the solution you obtain in each case. By comparing the colors you observe with the information in Table 25.1, estimate the pH of the solution to within a range of one pH unit, say 1 to 2, or 4 to 5. In making your estimate, note that the color of an indicator is most indicative of pH in the region where the indicator is changing color. Repeat the procedure with each of the following solutions: 0.10 M NaH2PO4

0.10 M HC2H3O2 0.10 M ZnSO4

Record the colors you observe and the pH range for each solution.


202

Experiment 25


Table 25.1 Useful pH Range (Approximate) Indicator

0

Methyl violet

1

2

yellow

Thymol blue

3

4

5

6

7

violet red

Methyl yellow

yellow red

Congo red

yellow violet

Bromcresol green

orange-red

yellow

blue

B. Measurement of the pH of Some Typical Solutions In the rest of this experiment we will use pH meters to find pH. Your instructor will show you how to operate your meter. The electrodes may be fragile, so use due caution when handling the electrode probe. See Appendix IV for a discussion of pH meters. Using a 25-mL sample in a 150-mL beaker, measure and record the pH of a 0.10 M solution of each of the following substances: NaCl Na2CO3

NaC2H3O2

NaHSO4

Rinse the electrode probe in distilled water between measurements. After you have completed a measurement, add a drop or two of bromcresol green to the solution and record the color you obtain. Some of the solutions are nearly neutral; others are acidic or basic. For each solution having a pH less than 6 or greater than 8, write a net ionic equation that explains qualitatively why the observed pH value is reasonable. Then write a rationale for the colors obtained with bromcresol green with these solutions.

C. Some Properties of Buffers On the lab bench we have 0.10 M stock solutions that can be used to make three different common buffer systems. These are HC2 H3O2 –– C2 H3O2 − acetic acid–acetate ion

NH 4 + –– NH3 ammonium ion–ammonia

HCO3− –– CO32− hydrogen carbonate–carbonate

The sources of the ions will be sodium and ammonium salts containing those ions. Select one of these buffer systems for your experiment. 1. Using a graduated cylinder, measure out 15 mL of the acid component of your buffer into a 100-mL beaker. The acid will be one of the following solutions: 0.10 M HC2H3O2, 0.10 M NH4Cl, or 0.10 M NaHCO3. Rinse out the graduated cylinder with distilled water and use it to add 15 mL of the conjugate base of your buffer. Measure the pH of your mixture and record it on the Data page. Calculate pKa for the acid. 2. Add 30 mL water to your buffer mixture, mix, and pour half of the resulting solution into another 100-mL beaker. Measure the pH of the diluted buffer. Calculate pKa once again. Add five drops of 0.10 M NaOH to the diluted buffer and measure the pH again. To the other half of the diluted buffer add 5 drops 0.10 M HCl, and again measure the pH. Record your results. 3. Make a buffer mixture containing 2 mL of the acid component and 20 mL of the solution containing the conjugate base. Mix, and measure the pH. Calculate a third value for pKa. To that solution add 3 mL 0.10 M NaOH. Measure the pH. Explain your results.


Experiment 25


203

4. Put 25 mL distilled water into a 100-mL beaker. Measure the pH. Add 5 drops 0.10 M HCl and measure the pH again. To that solution add 10 drops 0.10 M NaOH, mix, and measure the pH. 5. Select a pH different from any of those you observed in your experiments. Design a buffer which should have that pH by selecting appropriate volumes of your acidic and basic components. Make up the buffer and measure its pH.

D. Preparation of a Buffer from a Solution of a Weak Acid So far in this experiment the buffers that we used were made up from a weak acid and its conjugate base. Chemists faced with making a buffer would take a simpler approach. They would start with a solution of a weak acid with a pKa roughly equal to the pH of the buffer that was needed. To the acid they would slowly add an NaOH solution from a buret, stirring well, while at the same time measuring the pH of the solution. When they got to the desired pH, they would stop adding the NaOH. The buffer would be ready to use. In this part of the experiment you are to work alone. We will furnish you with an 0.50 M solution of a weak acid with a known pKa, and the pH of the buffer you will be asked to prepare. First, dilute the acid solution to 0.10 M by adding 10 mL of the acid to 40 mL water in a 100 mL beaker and stirring well. Using Equation 4b, calculate the ratio of [B−] to [HB] in the buffer to be prepared. Calculate how much 0.10 M NaOH you will have to add to 20 mL of your 0.10 M acid solution to produce the required ratio. This is easily done. If we add y mL of the NaOH, the value of [B−]/[HB} will become equal to y/(20 − y). Can you see why? Think about it, and you will soon see that is the case. Then complete the calculation and record the volume of NaOH that should be needed. Now do the experiment to check your prediction. Use the buret containing 0.10 M NaOH solution that has been set up by the pH meter. Record the volume before starting to add the base, noting the pH of the acid. Slowly add the NaOH to 20 mL of the acid, stirring well, and watching the pH as it slowly goes up. When you obtain a solution of the pH to be prepared, stop adding NaOH. Record the volume reading on the buret. Report the volume required to produce your buffer. Optional Experiment: Vitamin C (MM = 176 g) is a weak acid, called ascorbic acid. Study the buffering properties of Vitamin C using a solution made from a 500 mg tablet. Find Ka for Vitamin C and the pH range over which it might be useful as a buffer. When you are finished with the experiment, you may dilute the solutions with water and pour them down the drain.




Name ____________________________________

Section ________________________________

Experiment 25

Observations and Calculations: pH: Buffers and Their Properties A. Determination of pH by the Use of Acid-Base Indicators Color with 0.1 M Solution of Indicator

HCl

NaH2PO4

HC2H3O2

ZnSO4

Methyl violet

____________

____________

____________

____________

Thymol blue

____________

____________

____________

____________

Methyl yellow

____________

____________

____________

____________

Congo red

____________

____________

____________

____________

Bromcresol green

____________

____________

____________

____________

pH range

____________

____________

____________

____________

Circle the observation(s) for each solution that was most useful in estimating the pH range.

B. Measurement of the pH of Some Typical Solutions Record the pH and the color observed with bromcresol green for each of the 0.1 M solutions that were tested. NaCl

Na2CO3

NaC2H3O2

NaHSO4

pH

____________

____________

____________

____________

Color

____________

____________

____________

____________

For any two solutions having a pH less than 6 or greater than 8, write a net ionic equation to explain qualitatively why the solution has that pH. Solution ____________ Equation ______________________________________________________ Solution ____________ Equation ______________________________________________________ Explain why the color observed with bromcresol green for each of the four solutions is reasonable, given the pH.

C. Some Properties of Buffers Buffer system selected __________________ HB is __________________ (name the acid) 1. pH of buffer

____________ [H+] ____________ M pKa (by Eq. 4b) ____________

2. pH of diluted buffer

____________ [H+] ____________ M

pKa ____________

pH after addition of 5 drops NaOH

____________

pH after addition of 5 drops HCl

____________

Comment on your observations in Parts 1 and 2. (continued on following page) Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

206

Experiment 25


3. pH of buffer in which [HB]/[B−] = 0.10 pH after addition of excess NaOH

____________ pKa ____________ ____________

Explain your observations 4. pH of distilled water

____________

pH after addition of 5 drops HCl

____________

pH after addition of 5 drops NaOH

____________

Explain your observations 5. pH of buffer solution to be prepared Average value of pKa (as found in Parts 1, 2, and 3)

____________ ____________

[B−]/[HB] in buffer ____________ Volume 0.10 M NaB/Volume 0.10 M HB needed in buffer ____________ Volume 0.10 M NaB used ____________ mL

Volume 0.10 M HB used ____________ mL

pH of buffer you prepared ____________

D. Preparing a Buffer from a Solution of a Weak Acid pH of buffer to be prepared ____________ pKa of acid ____________ Ratio of [B−]/[HB] required in buffer

____________

Volume of 0.10 M NaOH calculated

____________ mL

pH of acid solution before titration

____________

Initial volume reading of NaOH

____________ mL

Final volume reading

____________ mL

Volume of NaOH actually required

____________ mL

How does that volume compare to your calculated value?


Name ____________________________________

Section ________________________________

Experiment 25

Advance Study Assignment: pH, Its Measurement and Applications 1. A solution of a weak acid was tested with the indicators used in this experiment. The colors observed were as follows: Methyl violet Thymol blue Methyl yellow

violet orange red

Congo red Bromcresol green

violet yellow

What is the approximate pH of the solution? ____________ 2. An aqueous solution of NH3 has a pH of 11.6. The ammonia (NH3) molecule is the conjugate base of the NH 4 + (called “ammonium”) ion. Write the net ionic equation for the reaction that makes an aqueous solution of NH3 basic. (Eq. 5.) 3. The pH of a 0.10 M HOBr solution is 4.8. a. What is [H+] in that solution? ____________ M b. What is [OBr −]? What is [HOBr]? (Where do the H+ and OBr − ions come from?) ____________ M; ____________ M c. What is the value of Ka for HOBr? What is the value of pKa? ____________

____________

4. Formic acid, HFor, has a Ka value equal to about 1.8 × 10−4. A student is asked to prepare a buffer having a pH of 3.90 from a solution of formic acid and a solution of sodium formate having the same molarity. How many milliliters of the NaFor solution should she add to 20. mL of the HFor solution to make the buffer? (See discussion of buffers.) ____________ mL 5. How many mL of 0.10 M NaOH should the student add to 20 mL 0.10 M HFor if she wished to prepare a buffer with a pH of 3.90, the same as in Problem 4? ____________ mL



Experiment 26 Determination of the Solubility Product of Ba(IO3)2

n several of the earlier experiments in this manual we used the concept of the solubility product in various applications. Every insoluble inorganic salt has an associated number, called its solubility product, that influences its behavior in mixtures where it is present as a solid. This number puts a condition on the concentrations of the ions in a mixture in which the salt is present. The equation expressing the condition follows from the chemical formula of the solid. For barium iodate the equation for the condition on the solubility product, Ksp, takes its form from the equation by which the solid dissolves:

I

Ba(IO3)2(s) = Ba2+(aq) + 2 IO3−(aq)

(1)

Ksp = [Ba2+] [IO3−]2

(2)

where

and the concentrations of the barium and iodate ions are in moles per liter in the solution in which solid barium iodate is present. Equation 2 essentially states that in a mixture of barium and iodate ions in equilibrium with solid barium iodate the product of the molarity of the Ba2+ ion times the molarity of the IO3− ion squared cannot vary, no matter how you change the ion concentrations in the system. If one goes up, the other most go down, keeping the ion product in Equation 2 exactly at its constant value. That’s rather amazing. For barium iodate, Ksp has a small value, less than 10−7. So, if one ion has a typical concentration, say 0.1 M, the other must have a very low concentration. If you add 0.1 M BaCl2 to a small volume of a solution of 0.05 M KIO3, you can be sure that a reaction will occur, in which Ba(IO3)2 precipitates, where the final solution will contain almost no iodate ion, and the product of the ions in Equation 2 has a fixed value. This behavior is seen in many analyses of practical importance. In Experiment 7, in which a solution of chloride ion is titrated with a solution of AgNO3, containing silver and nitrate ions, when the Ag+ ions first meet the Cl− ions, insoluble AgCl precipitates, since the product, Ksp, of their concentrations cannot be very large. As the silver nitrate is slowly added, AgCl precipitates until the Cl− ion is essentially gone, and the Ag+ ion concentration can finally go up. At the end point, silver dichlorofluorosceinate will form on the surface of the silver chloride particles. In this experiment we will be examining the solubility behavior of Ba(IO3)2 in various mixtures of Ba(NO3)2 and KIO3 solutions, for the purpose of ultimately measuring the Ksp of barium iodate and observing whether it indeed remains constant.

Experimental Procedure This is an experiment that can give excellent results if it is done carefully. It is probably the most difficult experiment in the manual to do properly. You will be working with small amounts of the chemical reactants, so small errors will be important.

Preparing the Reaction Mixtures From the stock solutions that are available, measure out about 35 mL of 0.0200 M Ba(NO3)2 into a small dry beaker. To a second small beaker add about 20 mL 0.0350 M KIO3. Use the labeled graduated cylinders next to the reagent bottles for measuring out these solutions. In the experiment we will precipitate solid Ba(IO3)2 under several different conditions, with varying proportions of the reagents you will be using. We will analyze the solutions remaining after the precipitation for Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

210

Experiment 26

Determination of the Solubility Product of Ba(lO3 )2

their iodate ion concentrations at equilibrium, and from those values and the way you made the mixtures, we can find the concentrations of barium ion, and the value of Ksp for barium iodate. Label five regular (18 × 150 mm) dry test tubes, either with labels or by noting their location in your test tube rack. To Test Tube 1, carefully pipet 1.00 mL 0.0350 M KIO3. Add 2, 3, 4, and 5 mL of that reagent to Test Tubes 2, 3, 4, and 5, respectively. You may use your 10-mL graduated pipet for these additions. (The easiest way to do this is to fill the pipet to the 0.00 level. With the pipet in Test Tube 1, let the level fall to 1.00. In Tube 2, let the level fall to 3.00, thereby adding 2 mL to the Tube. In Tube 3, let the level fall to 6.00, and in Tube 4 let it go to 10.00. Refill the pipet to the 5.00 level and in Tube 5 let it fall to 10.00.) Then, to each of the test tubes add 5.00 mL of 0.0200 M Ba(NO3)2, using your 5-mL pipet. Finally, pipet 6, 5, 4, 3, and 2 mL of distilled water into Test Tubes 1 through 5, respectively. The final volume in each tube should be 12.00 mL. The compositions of the mixtures are summarized in Table 26.1. Table 26.1 Volumes of Reagents Used in Precipitating Ba(IO3)2 (mL) Test Tube 1 2 3 4 5

0.0200 M Ba(NO3)2

0.0350 M KIO3

H2O

5.00 5.00 5.00 5.00 5.00

1.00 2.00 3.00 4.00 5.00

6.00 5.00 4.00 3.00 2.00

With a stirring rod, stir each solution for 30 seconds, up and down as well as swirling, wiping the rod on a paper towel before proceeding to the next solution. Touch the wall of the tube occasionally as you stir. Look at each tube. In most of them you should see a crystalline white precipitate, which slowly settles out. It will be most noticeable in the higher number tubes, but it should be present as a light cloudiness in Test Tubes 1 and 2. There will only be a tiny amount of precipitate, only a few mg, so the amount is not impressive. It is most easily seen if you look through the tube toward the light, where you may observe tiny particles, very slowly settling out. The crystallization occurs slowly, so give it time. Put a cork stopper in each of the tubes in which there is a precipitate, and slowly rotate the tubes so that the liquid goes from end to end twenty times or so to ensure thorough mixing and help bring the system to equilibrium. (For any tubes in which you observe only a clear liquid, and no solid, repeat the stirring operation, scratching the walls of the tube as you stir, which may get things started. If that still doesn’t do the job, touch your stirring rod to the sample of pure solid barium iodate that we have in the lab and put the rod in the reluctant liquid. Insert a cork stopper in the tube and shake with vigor. Sooner or later you will get the cloudiness to form.) Repeat the rotation step, turning each tube end to end to do what you can to make the liquid homogeneous and in equilibrium with the precipitate. You can’t overstir, so be patient. Give yourself five minutes with this. It will be well spent. Finally, let the test tubes stand for 15 minutes to let the solid precipitate particles settle out completely, leaving a clear solution above the solids. While you are waiting for the precipitates to settle out, you can begin the calculations you will need later. Fill in as many blanks as you can on the Data and Calculation page, or on your Excel sheet. From the way you made your mixtures, you can determine the numbers of moles of Ba2+ and IO3− that were initially present in each tube. Once you have completed the analysis of the solutions at equilibrium you can finish the calculations of Ksp.

Preparing the Equilibrium Mixtures for Analysis The mixtures in your test tubes at this point should contain essentially clear solutions above the precipitates that formed. You now need to remove an aliquot from each of those solutions, which should now be at equilibrium, without disturbing the solid at the bottom of the test tubes. This is the most difficult part of the experiment, and must be done very carefully. It is done with a 5-mL pipet. First, rinse and drain that pipet with distilled water. Then blow out the water remaining in the tip. Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Experiment 26

Determination of the Solubility Product of Ba(lO3 )2

211

Clamp Test Tube 1 on a ring stand, so that you can work on it without it being able to move. Immerse the 5-mL pipet about half-way down the clear solution, and hold it there with your hand. Compress a suction bulb slightly, and attach it gently to the top of the pipet. Slowly release the pressure on the bulb, and draw the solution into the pipet until it is near the top of the pipet. Remove the bulb and quickly put your finger on the top of the pipet to hold the liquid in place. Let liquid flow back into the test tube, slowly, slowly, until the level is at the marking line on the pipet. Remove the pipet from the test tube and let the contents flow into a clean dry test tube. This operation is easier said than done. If something goes wrong while you are drawing the liquid into the pipet, remove the bulb and let the liquid go back into the test tube. After things settle down, you can try again, until you get it right. In order to increase your chances of accomplishing this transfer properly, practice doing it with water in a clamped test tube. After about five successful practice runs, you should be able to make the transfer where it counts. So, go ahead and do it, for each of the five test tubes, using five clean dry test tubes to receive the 5-ml aliquots.

Analyzing the Equilibrium Solutions Having separated 5 mL of each of the five equilibrium solutions, it is now possible to analyze for the concentration of IO3− ion in those solutions. We do it by reducing iodate ion to I2, which is colored and can be analyzed spectroscopically. The reaction we will use is: IO3−(aq) + 5 I−(aq) + 6 H+(aq) → 3 I2(aq) + 3 H2O*

(3)

To each of the test tubes add, by pipet, 1.00 mL 1 M KI and 1.00 mL 1 M HCl, and 3.00 mL distilled water, bringing the volume of the five solutions to 10.00 mL. During this step the color of the solution will become orange as the reaction proceeds. Stir each mixture well with a stirring rod, drying the rod on a paper towel before you go on to the next tube. Go to the Spectronic 20 in the lab. The wavelength setting should be 500 nm. If you have not used the Spec 20 before, read up in Appendix IV on how to use it to measure light absorption. Pick up a Spectronic 20 tube, and if it is not dry, shake it to remove as much liquid as you can, and then add about a milliliter of the orange liquid from Tube 1, shake to mix it with the old liquid, and then pour that liquid into a waste beaker. Rinse, and drain again with another 1-mL portion of your solution. Then fill the Spec 20 tube up to the level of the label with the liquid in Tube 1. Set the zero and 100% readings if they have not yet been adjusted; then put your tube in the tube holder and read the absorbance of the solution. The concentration of the IO3− ion in the equilibrium 12-mL mixture will equal the value on the graph in the lab. (This is drawn to take account of the dilution you made.) Record the absorbance and concentration for Tube 1. Repeat these measurements for the other four tubes you worked with, and record your values for those solutions. Pour all of the solutions and solids in the ten test tubes you used in the waste crock. Rinse out the tubes with distilled water.

Processing the Data From the volume and concentration of the KIO3 you used in making the five mixtures of reagents, you have probably already found the number of moles of iodate ion originally in each mixture. From the concentration of the IO3− ion that we measured for each of the equilibrium solutions we calculate the number of moles of that ion in those solutions, with its 12-mL volume. The difference between the original number of moles and the equilibrium number of moles must be the number of moles of iodate ion in the precipitate at the bottom of the tube. The number of moles of barium in the precipitate must equal half of that value, given the formula of barium iodate, Ba(IO3)2. Why? The number of moles of barium ion that remain in the solution will equal the difference between the number in the original mixture and the number you found were in the precipitate in each of the mixtures. Knowing the volume of the equilibrium solution, 12.00 mL, you can calculate the concentration of the Ba2+ ion in each mixture. Record the results of these calculations on the Data and Calculations page or on your Excel sheet. Finally, using the concentrations you have recorded, find the values of Ksp for each of the five mixtures. Calculate the mean value of Ksp and the standard deviation. (See Appendix VIII.) *The actual iodine species is I3−, but I2 is simpler so we use it. Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.


Name __________________________

Section __________________________

Experiment 26

Data and Calculations: The Solubility Product of Ba(IO3)2 The calculations are similar to those in Experiment 23, and are outlined in the Experimental Procedure section. To use Excel, set up a table like the one below. Test Tube No.

1

2

3

4

5

mL 0.0350 M KIO3

___________

___________

___________

___________

___________

mL 0.0200 M Ba(NO3)2

___________

___________

___________

___________

___________

Total volume in mL

___________

___________

___________

___________

___________

Absorbance of soln.

___________

___________

___________

___________

___________

[IO3−] in soln.

___________

___________

___________

___________

___________

Processing the Data (all data × 10–5 except for Ksp values) Initial no. moles Ba2+

___________

___________

___________

___________

___________

Initial no. moles IO3−

___________

___________

___________

___________

___________

No. moles IO3− in equil. soln. (12 mL)

___________

___________

___________

___________

___________

No. moles IO3− precipitated

___________

___________

___________

___________

___________

No. moles Ba2+ precipitated

___________

___________

___________

___________

___________

No. moles Ba2+ in equil. soln. (12 mL)

___________

___________

___________

___________

___________

[Ba2+] in equil. soln.

___________

___________

___________

___________

___________

Ksp for Ba(IO3)2

___________

___________

___________

___________

___________

Mean value of Ksp _____________________

Standard deviation _____________________

How many mg of solid barium iodate were there in the precipitate in Test Tube 1? MM Ba(IO3)2 = 487.14 __________ mg Does [Ba2+] vary as expected in the experiment? Explain. What is your best value of the solubility of Ba(IO3)2 in water? Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.


Name __________________________

Section __________________________

Experiment 26

Advance Study Assignment: Determination of the Solubility Product of Ba(IO3)2 1.

State in words what is meant by the solubility product equation for Ba(IO3)2: Ksp = [Ba2+] [IO3−]2

2.

If 10 mL 0.10 M Ba(NO3)2 is mixed with 10 mL 0.10 M KIO3, a precipitate forms. Which ion will still be present at appreciable concentration in the equilibrium mixture if Ksp for barium iodate is very small? Indicate your reasoning. What would that concentration be?

______ ___________ moles/L 3.

Lead chloride, PbCl2, is moderately soluble, and has a Ksp equal to about 1.7 × 10−5. a. What would be the solubility of lead chloride in water?

___________ moles/L b. What would be its solubility in 0.10 M NaCl?

___________ moles/L c. The difference is caused by what is called the common ion effect. State in words what the common ion effect predicts.



Experiment 27 Relative Stabilities of Complex Ions and Precipitates Prepared from Solutions of Copper(II)

n aqueous solution, typical cations, particularly those produced from atoms of the transition metals, do not exist as free ions but rather consist of the metal ion in combination with some water molecules. Such cations are called complex ions. The water molecules, usually two, four, or six in number, are bound chemically to the metallic cation, but often rather loosely, with the electrons in the chemical bonds being furnished by one of the unshared electron pairs from the oxygen atoms in the H2O molecules. Copper ion in aqueous solution may exist as Cu(H 2O)4 2+ , with the water molecules arranged in a square around the metal ion at the center. If a hydrated cation such as Cu(H 2O)4 2+ is mixed with other species that can, like water, form coordinate covalent bonds with Cu2+, those species, called ligands, may displace one or more H2O molecules and form other complex ions containing the new ligands. For instance, NH3, a reasonably good coordinating species, may replace H2O from the hydrated copper ion, Cu(H 2O)4 2+ , to form Cu(H 2O)3 NH32+ , Cu(H 2O)2 (NH3 )2 2+ , Cu(H 2O)(NH3 )32+ , or Cu(NH3 )4 2+ . At moderate concentrations of NH3, essentially all the H2O molecules around the copper ion are replaced by NH3 molecules, forming the copper ammonia complex ion. Coordinating ligands differ in their tendencies to form bonds with metallic cations, so that in a solution containing a given cation and several possible ligands, an equilibrium will develop in which most of the cations are coordinated with those ligands with which they form the most stable bonds. There are many kinds of ligands, but they all share the common property that they possess an unshared pair of electrons that they can donate to form a coordinate covalent bond with a metal ion. In addition to H2O and NH3, other uncharged coordinating species include CO and ethylenediamine; some common anions that can form complexes include 2− OH−, Cl−, CN−, SCN−, and S2O3 . As you know, when solutions containing metallic cations are mixed with other solutions containing ions, precipitates are sometimes formed. When a solution of 0.1 M copper nitrate is mixed with a little 1 M NH3 solution, a precipitate forms and then dissolves in excess ammonia. The formation of the precipitate helps us to understand what is occurring as NH3 is added. The precipitate is hydrous copper hydroxide, formed by reaction of the hydrated copper ion with the small amount of hydroxide ion present in the NH3 solution. The fact that this reaction occurs means that even at very low OH− ion concentration Cu(OH)2(H2O)2(s) is a more stable species than Cu(H 2O)4 2+ ion. Addition of more NH3 causes the solid to redissolve. The copper species then in solution cannot be the 2+ hydrated copper ion. (Why?) It must be some other complex ion and is, indeed, the Cu(NH3 )4 ion. The implication of this reaction is that the Cu(NH3 )4 2+ ion is also more stable in NH3 solution than is the hydrated copper ion. To deduce in addition that the copper ammonia complex ion is also more stable in general than Cu(OH)2(H2O)2(s) is not warranted, since under the conditions in the solution [NH3] is much larger than [OH−], and given a higher concentration of hydroxide ion, the solid hydrous copper hydroxide might possibly precipitate even in the presence of substantial concentrations of NH3. To resolve this question, you might proceed to add a little 1 M NaOH solution to the solution containing the Cu(NH3 )4 2+ ion. If you do this you will find that Cu(OH)2(H2O)2(s) does indeed precipitate. We can conclude from these observations that Cu(OH)2(H2O)2(s) is more stable than Cu(NH3 )4 2+ in solutions in which the ligand concentrations (OH− and NH3) are roughly equal. The copper species that will be present in a system depends, as we have just seen, on the conditions in the system. We cannot say in general that one species will be more stable than another; the stability of a given species depends in large measure on the kinds and concentrations of other species that are also present with it.

I


218

Experiment 27

Relative Stabilities of Complex Ions and Precipitates Prepared from Solutions of Copper(II)

Another way of looking at the matter of stability is through equilibrium theory. Each of the copper species we have mentioned can be formed in a reaction between the hydrated copper ion and a complexing or precipitating ligand; each reaction will have an associated equilibrium constant, which we might call a formation constant for that species. The pertinent formation reactions and their constants for the copper species we have been considering are listed here: Cu(H 2O)4 2+ (aq) + 4 NH3(aq)

Cu(NH3 )4 2+ (aq) + 4 H2O

K1 = 5 × 1012

(1)

Cu(H 2O)4 2+ (aq) + 2 OH−(aq)

Cu(OH)2(H2O)2(s) + 2 H2O

K2 = 2 × 1019

(2)

The formation constants for these reactions do not involve [H2O] terms, which are essentially constant in aqueous systems and are included in the magnitude of K in each case. The large size of each formation constant indicates that the tendency for the hydrated copper ion to react with the ligands listed is very high. In terms of these data, let us compare the stability of the Cu(NH3 )4 2+ complex ion with that of solid Cu(OH)2(H2O)2. This is most readily done by considering the reaction: Cu(NH3 )4 2+ (aq) + 2 OH−(aq) + 2 H2O

Cu(OH)2(H2O)2(s) + 4 NH3(aq)

(3)

We can find the value of the equilibrium constant for this reaction by noting that it is the sum of Reaction 2 and the reverse of Reaction 1. By the Law of Multiple Equilibrium, K for Reaction 3 is given by the equation K =

[ NH3 ]4 2 × 1019 K2 = = 4 × 106 = 12 K1 5 × 10 [Cu( NH3 )4 2+ ][OH − ]2

(4)

From the expression in Equation 4 we can calculate that in a solution in which the NH3 and OH− ion concentrations are both about 1 M. [Cu( NH3 )4 2+ ] =

1 = 2.5 × 10 −7 M 4 × 106

(5)

Since the concentration of the copper ammonia complex ion is very, very low, any copper(II) in the system will exist as the solid hydroxide. In other words, the solid hydroxide is more stable under such conditions than the ammonia complex ion. But that is exactly what we observed when we treated the hydrated copper ion with ammonia and then with an equivalent amount of hydroxide ion. Starting now from the experimental behavior of the copper ion, we can conclude that since the solid hydroxide is the species that exists when copper ion is exposed to equal concentrations of ammonia and hydroxide ion, the hydroxide is more stable under those conditions, and the equilibrium constant for the formation of the hydroxide is larger than the constant for the formation of the ammonia complex. By determining, then, which species is present when a cation is in the presence of equal ligand concentrations, we can speak meaningfully of stability under such conditions and can rank the formation constants for the possible complex ions, and indeed for precipitates, in order of their increasing magnitudes. In this experiment you will carry out formation reactions for a group of complex ions and precipitates involving the Cu2+ ion. You can make these species by mixing a solution of Cu(NO3)2 with solutions containing NH3 or anions, which may form either precipitates or complex ions by reaction with Cu(H 2O)4 2+ , the cation present in aqueous solutions of copper(II) nitrate. By examining whether the precipitates or complex ions formed by the reaction of hydrated copper(II) ion with a given species can, on addition of a second ligand, be dissolved or transformed to another species, you will be able to rank the relative stabilities of the precipitates and complex ions made from Cu2+ with respect to one another, and thus rank the equilibrium formation constants for each species in order of increasing magnitude. The species to be reacted with Cu2+ ion in aqueous solution are NH3, Cl−, OH−, C2O 4 2− , S2−, NO2 − , and PO 43− . In each case the test for relative stability will be made in the presence of essentially equal concentrations of the two ligands. When you have completed your ranking of the known species you will test an unknown species and incorporate it into your list.


Experiment 27


219

Experimental Procedure Obtain from the stockroom an unknown and eight small test tubes. Add about 1 mL 0.1 M Cu(NO3)2 solution to each of the test tubes ( 1 2 -inch depth). To one of the test tubes add about 1 mL 1 M NH3, drop by drop. Note whether a precipitate forms initially, and if it dissolves in excess NH3. Shake the tube sideways to mix the reagents. In the NH3-NH3 space in the table (Data page), write a P in the upper left-hand corner if a precipitate is formed initially. In the rest of the space, write the formula and the color of the species that was present with an excess of NH3. If a solution is present, the copper ion will be in a complex. Cu(II) will always have a coordination number of 4, so the formula with NH3 would be Cu(NH3 )4 2+ . If a precipitate is present, it will be neutral, and in the case of NH3 it would be a hydroxide with the formula Cu(OH)2. (There should in principle be two H2O molecules in the formula, but they are usually omitted.) Add 1 mL 1 M NH3 to the rest of the test tubes. Now you will test the stability of the species present in excess NH3 relative to those which might be present with other precipitating or coordinating species. Add, drop by drop, 1 mL of a 1 M solution of each of the anions in the horizontal row of the table to the test tubes you have just prepared, one solution to a test tube. Note any changes that occur in the appropriate spaces in the table. A change in color or the formation of a precipitate implies that a reaction has occurred between the added ligand or precipitating anion and the species originally present. As before, put a P in the upper left-hand corner if a precipitate initially forms on addition of the anion solution. In the rest of the space, write the formula and color of the species present when the anion is in excess. That is the species that is stable in the presence of equal concentrations of NH3 and the added anion. Again, in complex ions, Cu(II) will normally be attached to four ligands; copper(II) precipitates will be neutral. If a new species forms on addition of the second reagent, its formula should be given. If no change occurs, the original species is more stable, so put its formula in that space. Repeat the above series of experiments, using 1 M Cl− as the species originally added to the Cu(NO3)2 solution. In each case record the color of any precipitates or solutions formed on addition of the reagents in the horizontal row, and the formula of the species stable when an excess of both Cl− ion and the added species is present in the solution. Since these reactions are reversible, it is not necessary to retest Cl− with NH3, since the same results would be obtained as when the NH3 solution was tested with Cl− solution. Repeat the series of experiments for each of the anions in the vertical row in the table, omitting those tests where decisions as to relative stabilities are already clear. Where both ligands produce precipitates it will be helpful to check the effect of the addition of the other ligand to those precipitates. When complete, your table should have at least 28 entries. Examine your table and decide on the relative stabilities of all species you observed to be present in all the spaces of the table. There should be seven such species; rank them as best you can in order of increasing stability. There is only one correct ranking, and you should be prepared to defend your choices. Although we did not in general prepare the species by direct reaction of Cu(H 2O)4 2+ with the added ligand or precipitating anion, the equilibrium formation constants for those species for the direct reactions will have magnitudes that increase in the same order as the relative stabilities of the species you have established. When you are satisfied that your ranking order is correct, carry out the necessary tests on your unknown to determine its proper position in the list. Your unknown may be one of the Cu(II) species you have already observed, or it may be a different species, present in excess of its ligand or precipitating anion. If your unknown contains a precipitate, shake it well before using a portion of it to make a test. Dispose of all reaction products in the waste crock provided or as otherwise directed by your instructor.

Alternate Procedure Using Microscale Approach

Optional

From the stockroom obtain an unknown and two plastic well plates (4 × 6). Align the plates into a 6 × 8 well configuration, with six wells across the top. To each of the six wells in the top row of the plate add six drops of 0.1 M Cu(NO3)2. Use a Beral pipet for this and all other additions of reagents. To the six wells add six drops of 1 M solutions containing the species in the horizontal row at the


220

Experiment 27


top of the Table of Observations; one species to a well, starting with 1 M NH3 in the first well and ending with 1 M NaNO2 in the sixth (we will omit S2− and come back to it later). Mix the reagents by moving the well plate back and forth on the top of the lab bench. You will notice a difference in the appearance of the mixtures in each of the six wells. Each contains the species that is stable when Cu(II) is mixed with one of the reagents. Report the color and the formula of those six species along the diagonal blocks in the Table, using the directions in the first two paragraphs in the ordinary procedure. The contents of these wells will serve as a reference during the rest of the experiment. To establish the relative stabilities of the six species, we need to mix them with each of the other species in some systematic way and see what happens. We will first test them against NH3. To do this, in the second row of wells make the same mixtures as you did in the first row. Then, add six drops of 1 M NH3 to the second through sixth wells in that row (we do not need to test NH3 with NH3, since we really have already done that). Mix the reagents, and note any changes that occurred on addition of NH3. In some of the wells, changes will occur, indicating a reaction to form a more stable species has occurred. Use the wells in the top row for comparison. Record your observations in the blocks in the first row of the Table, noting the color and formula of the stable Cu(II) species in each case. Now make stability tests for Cl−, using wells in the third row of the plate. This time start with the third well, since we already know how Cl− and NH3 stack up from the last set of tests. (It doesn’t make any difference if we change the order in which the reagents are added.) The four wells on the right end of the third row should be prepared in the same way as the corresponding wells in the first row. To each of those wells add six drops of 1 M NaCl, mix, note any changes, and report your observations as before in the blocks in the second row of the Table. Continue along these lines, making stability tests with 1 M NaOH in the bottom row of the top plate. Report your observations as before. Then, in the top three rows of the lower well plate, make the same sort of tests with 1 M K2C2O4, 1 M Na2HPO4, and 1 M KNO2, respectively. When you are done, you should have entries in all of the blocks in the upper right half of the Table except for those involving S2−. To establish the stability of the Cu(II) species with respect to the sulfide ion, S2−, add a drop or two of 0.1 M (NH4)2S to each of the wells in the top row of the top plate. Although the sulfide solution is dilute, its effect on the contents of the wells, and on the nose, should be clear. Fill in the last column of the Table. Given your entries in the Table, you should be able to decide on the relative stabilities of the seven Cu(II) species studied in this experiment. Repeat any tests that appear ambiguous, changing the order in which reagents are added if necessary. List the seven species, in order of increasing stability. Then carry out the necessary tests on your unknown to establish its position in the list. The unknown may be one of the Cu(II) species you have studied, or it may be different. If the unknown contains a precipitate, stir it up before making a test. Pour the contents of all of the wells into the waste crock unless directed otherwise by your instructor.



Name ____________________________________

Section ________________________________

Experiment 27

Observations and Conclusions: Relative Stabilities of Complex Ions and Precipitates Prepared from Solutions of Copper(II)

Unknown

S2−

Cu(OH)2

PO 43−

C2O 4 2−

OH−

Cl−

NH3

NH3

Cl−

OH−

C2O 4 2−

PO 43−

Cu(OH)2

S2−

Table of Observations

(continued on following page) Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

222

Experiment 27


Determination of Relative Stabilities In each row of the table you can compare the stabilities of species involving the reagent in the horizontal row with those of the species containing the reagent initially added. In the first row of the table, the copper(II)NH3 species can be seen to be more stable than some of the species obtained by addition of the other reagents, and less stable than others. Examining each row, make a list of all the complex ions and precipitates you have in the table in order of increasing stability and formation constant.

Lowest

Highest

_________________________________

Reasons _________________________________

_________________________________

_________________________________

_________________________________

_________________________________

_________________________________

_________________________________

_________________________________

_________________________________

_________________________________

_________________________________

_________________________________

_________________________________

Stability of Unknown Indicate the position your unknown would occupy in the above list. Reasons:

Unknown no.____________


Name ____________________________________

Section ________________________________

Experiment 27

Advance Study Assignment: Relative Stabilities of Complex Ions and Precipitates Prepared from Solutions of Copper(II) 1. In testing the relative stabilities of Cu(II) species using a well plate, a student adds 6 drops 1 M NH3 to 6 drops 0.1 M Cu(NO3)2. He observes that a blue precipitate initially forms, but that in excess NH3 the precipitate dissolves and the solution turns blue. Addition of 6 drops 1 M NaOH to the dark-blue solution results in the formation of a blue precipitate. a. What is the formula of Cu(II) species in the dark-blue solution?

b. What is the formula of the blue precipitate present after addition of 1 M NaOH?

c. Which species is more stable in equal concentrations of NH3 and OH−, the one in Part a or the one in Part b? ____________ 2. Given the following two reactions and their equilibrium constants: Cu(H 2O)4 2+ (aq) + 4 NH3(aq) Cu(H 2O)4 2+ (aq) + CO32−

Cu(NH3 )4 2+ (aq) + 4 H2O K1 = 5 × 1012 CuCO3(s) + 4 H2O

K2 = 7 × 109

a. Evaluate the equilibrium constant for the reaction (see discussion): CuCO3(s) + 4 NH3(aq)

Cu(NH3 )4 2+ (aq) + CO32− (aq) ____________

b. If 1 M NH3 were added to some solid CuCO3 in a test tube containing 1 M Na2CO3, what, if anything, would happen? Explain your reasoning.



Experiment 28 Determination of the Hardness of Water

ne of the factors that establishes the quality of a water supply is its degree of hardness. The hardness of water is defined in terms of its content of calcium and magnesium ions. Since the analysis does not distinguish between Ca2+ and Mg2+, and since most hardness is caused by carbonate deposits in the earth, hardness is usually reported as total parts per million calcium carbonate by weight. A water supply with a hardness of 100 parts per million would contain the equivalent of 100 grams of CaCO3 in 1 million grams of water or 0.1 gram in one liter of water. In the days when soap was more commonly used for washing clothes, and when people bathed in tubs instead of using showers, water hardness was more often directly observed than it is now, since Ca2+ and Mg2+ form insoluble salts with soaps and make a scum that sticks to clothes or to the bathtub. Detergents have the distinct advantage of being effective in hard water, and this is really what allowed them to displace soaps for laundry purposes. Water hardness can be readily determined by titration with the chelating agent EDTA (ethylenediaminetetraacetic acid). This reagent is a weak acid that can lose four protons on complete neutralization; its structural formula is HOOC CH2 CH2 COOH

O

N HOOC

CH2

CH2

CH2

N CH2

COOH

The four acid sites and the two nitrogen atoms all contain unshared electron pairs, so that a single EDTA ion can form a complex with up to six sites on a given cation. The complex is typically quite stable, and the conditions of its formation can ordinarily be controlled so that it contains EDTA and the metal ion in a 1:1 mole ratio. In a titration to establish the concentration of a metal ion, the EDTA which is added combines quantitatively with the cation to form the complex. The end point occurs when essentially all of the cation has reacted. In this experiment we will standardize a solution of EDTA by titration against a standard solution made from calcium carbonate, CaCO3. We will then use the EDTA solution to determine the hardness of an unknown water sample. Since both EDTA and Ca2+ are colorless, it is necessary to use a rather special indicator to detect the end point of the titration. The indicator we will use is called Eriochrome Black T, which forms a rather stable wine-red complex, MgIn−, with the magnesium ion. A tiny amount of this complex will be present in the solution during the titration. As EDTA is added, it will complex free Ca2+ and Mg2+ ions, leaving the MgIn− complex alone until essentially all of the calcium and magnesium have been converted to chelates. At this point EDTA concentration will increase sufficiently to displace Mg2+ from the indicator complex; the indicator reverts to an acid form, which is sky blue, and this establishes the end point of the titration. The titration is carried out at a pH of 10, in an NH3 NH4+ buffer, which keeps the EDTA (H4Y) mainly in the form, HY3−, where it complexes the Group 2 ions very well but does not tend to react as readily with other cations such as Fe3+ that might be present as impurities in the water. Taking H4Y and H3In as the formulas for EDTA and Eriochrome Black T, respectively, the equations for the reactions which occur during the titration are: (main reaction) HY3−(aq) + Ca2+(aq) → CaY2−(aq) + H+(aq) (same for Mg2+) (at end point) HY3−(aq) + MgIn−(aq) → MgY2−(aq) + HIn2−(aq) wine red sky blue Since the indicator requires a trace of Mg2+ to operate properly, we will add a little magnesium ion to each solution and titrate it as a blank.


226

Experiment 28

Determination of the Hardness of Water

Experimental Procedure Obtain a 50-mL buret, a 250-mL volumetric flask, and 25- and 50-mL pipets from the stockroom. Put about a half gram of calcium carbonate in a small 50-mL beaker and weigh the beaker and contents on the analytical balance. Using a spatula, transfer about 0.4 g of the carbonate to a 250-mL beaker and weigh again, determining the mass of the CaCO3 sample by difference. Add 25 mL of distilled water to the large beaker and then, slowly, about 40 drops of 6 M HCl. Cover the beaker with a watch glass and allow the reaction to proceed until all of the solid carbonate has dissolved. Rinse the walls of the beaker down with distilled water from your wash bottle and heat the solution until it just begins to boil. (Be sure not to be confused by the evolution of CO2 that occurs with the boiling.) Add 50 mL of distilled water to the beaker and carefully transfer the solution, using a stirring rod as a pathway, to the volumetric flask. Rinse the beaker several times with small portions of distilled water and transfer each portion to the flask. All of the Ca2+ originally in the beaker should then be in the volumetric flask. Fill the volumetric flask to the horizontal mark with distilled water, adding the last few mL a drop at a time with your wash bottle. Stopper the flask and mix the solution thoroughly by inverting the flask at least 20 times over a period of several minutes. Clean your buret thoroughly and draw about 200 mL of the stock EDTA solution from the carboy into a dry 250-mL Erlenmeyer flask. Rinse the buret with a few mL of the solution at least three times. Drain through the stopcock and then fill the buret with the EDTA solution. Determine a blank by adding 25 mL distilled water and 5 mL of the pH 10 buffer to a 250-mL Erlenmeyer flask. Add a small amount of solid Eriochrome Black T indicator mixture from the stock bottle. You need only a small portion, about 25 mg, just enough to cover the end of a small spatula. The solution should turn blue; if the color is weak, add a bit more indicator. Add 15 drops 0.03 M MgCl2, which should contain enough Mg2+ to turn the solution wine red. Read the buret to 0.02 mL and add EDTA to the solution until the last tinge of purple just disappears. The color change is rather slow, so titrate slowly near the end point. Only a few mL will be needed to titrate the blank. Read the buret again to determine the volume required for the blank. This volume must be subtracted from the total EDTA volume used in each titration. Save the solution as a reference for the end point in all your titrations. Pipet three 25-mL portions of the Ca2+ solution in the volumetric flask into clean 250-mL Erlenmeyer flasks. To each flask add 5 mL of the pH 10 buffer, a small amount of indicator, and 15 drops of 0.03 M MgCl2. Titrate the solution in one of the flasks until its color matches that of your reference solution; the end point is a reasonably good one, and you should be able to hit it within a few drops if you are careful. Read the buret. Refill the buret, read it, and titrate the second solution, then the third. Your instructor will furnish you a sample of water for hardness analysis. Since the concentration of Ca2+ is probably lower than that in the standard calcium solution you prepared, pipet 50 mL of the water sample for each titration. As before, add some indicator, 5 mL of pH 10 buffer, and 15 drops of 0.03 M MgCl2 before titrating. Carry out as many titrations as necessary to obtain two volumes of EDTA that agree within about 3%. If the volume of EDTA required in the first titration is low due to the fact that the water is not very hard, increase the volume of the water sample so that in succeeding titrations, it takes at least 20 mL of EDTA to reach the end point. Optional Experiment: Find the mass percent calcium carbonate in a Tums tablet or a sample of limestone. DISPOSAL OF REACTION PRODUCTS.

The chemical waste from this experiment may be diluted

with water and poured down the sink.


Name ____________________________________

Section ________________________________

Experiment 28

Data and Calculations: Determination of the Hardness of Water

____________ g

Volume Ca2+ solution prepared

____________ mL

____________ g

Molarity of Ca2+

____________ M

Mass of CaCO3 sample

____________ g

Moles Ca2+ in each aliquot titrated

____________ moles

Number of moles CaCO3 in sample (Formula mass = 100.1)

____________ moles

Mass of beaker plus CaCO3

Mass of beaker less sample

Standardization of EDTA Solution Determination of blank: Initial buret reading ____________ mL

Titration

Final buret reading ____________ mL

I

Volume of blank ____________ mL

II

III

Initial buret reading

____________ mL

____________ mL

____________ mL

Final buret reading

____________ mL

____________ mL

____________ mL

Volume of EDTA

____________ mL

____________ mL

____________ mL

Volume EDTA used to titrate blank

____________ mL

____________ mL

____________ mL

Volume EDTA used to titrate Ca2+

____________ mL

____________ mL

____________ mL

Average volume of EDTA required to titrate Ca2+ Molarity of EDTA =

no. moles Ca 2 + in aliquot × 1000 = average volume EDTA required (mL)

____________ mL ____________ M (continued on following page)


228

Experiment 28


Determination of Water Hardness Titration

I

II

III

Volume of water used

____________ mL

____________ mL

____________ mL


____________ mL

____________ mL

____________ mL

Final buret reading

____________ mL

____________ mL

____________ mL

Volume of EDTA

____________ mL

____________ mL

____________ mL

Volume of EDTA used to titrate blank

____________ mL

____________ mL

____________ mL

Volume of EDTA required to titrate water

____________ mL

____________ mL

____________ mL

Average volume EDTA per liter of water

____________ mL

No. moles EDTA per liter water ____________ = No. moles CaCO3 per liter water No. grams CaCO3 per liter water____________ g

Water hardness (1 ppm = 1 mg/liter)____________ ppm CaCO3

Unknown no. ____________


Name ____________________________________

Section ________________________________

Experiment 28

Advance Study Assignment: Determination of the Hardness of Water 1. A 0.4243-g sample of CaCO3 is dissolved in 12 M HCl and the resulting solution is diluted to 250.0 mL in a volumetric flask. a. How many moles of CaCO3 are used (formula mass = 100.1)?

____________ moles b. What is the molarity of the Ca2+ in the 250 mL of solution?

____________ M 2+

c. How many moles of Ca are in a 25.00-mL aliquot of the solution in 1b?

____________ moles

2. 25.00-mL aliquots of the solution from Problem 1 are titrated with EDTA to the Eriochrome Black T end point. A blank containing a small measured amount of Mg2+ requires 3.21 mL of the EDTA to reach the end point. An aliquot to which the same amount of Mg2+ is added requires 24.95 mL of the EDTA to reach the end point. a. How many milliliters of EDTA are needed to titrate the Ca2+ ion in the aliquot?

____________ mL b. How many moles of EDTA are there in the volume obtained in Part a?

____________ moles c. What is the molarity of the EDTA solution?

____________ M


230

Experiment 28


3. A 100-mL sample of hard water is titrated with the EDTA solution in Problem 2. The same amount of Mg2+ is added as previously, and the volume of EDTA required is 31.84 mL. a. What volume of EDTA is used in titrating the Ca2+ in the hard water?

____________ mL b. How many moles of EDTA are there in that volume?

____________ moles c. How many moles of Ca2+ are there in the 100 mL of water?

____________ moles d. If the Ca2+ comes from CaCO3, how many moles of CaCO3 are there in one liter of the water? How many grams of CaCO3 are present per liter of the water?

____________ moles ____________ g e. If 1 ppm CaCO3 = 1 mg per liter, what is the water hardness in ppm CaCO3?

____________ ppm CaCO3


Experiment 29 Synthesis and Analysis of a Coordination Compound

ome of the most interesting research in inorganic chemistry has involved the preparation and properties of coordination compounds. These compounds, sometimes called complexes, are typically salts that contain complex ions. A complex ion contains a central metal atom to which are bonded small polar molecules or anions, called ligands. The bonding in the complex is through coordinate covalent bonds, bonds in which the electrons are all furnished by the ligands. Some coordination compounds can be prepared stoichiometrically pure. That is, the atoms in the compound are present in the exact ratio given by the formula. Many supposedly pure compounds do not meet this criterion, since they contain variable amounts of water, either in hydrates or adsorbed. In this experiment we will be making a compound that is indeed stoichiometrically pure. It is not a hydrate nor does it tend to adsorb water. The compound we will be preparing contains cobalt ion, ammonia, and chloride ions. Its formula is of the form Cox(NH3)yClz. Once you have made the compound, you may, depending on the time available, analyze it for its content of one or more of these species and so determine x, y, and/or z. In many reactions involving complex ion formation the rate of reaction is very fast, so that the thermodynamically stable form of the complex is the one produced. By changing ligand concentrations, one can quickly convert from one complex to another. In Experiment 27, in which many Cu(II) complexes are prepared, all of the complex ions undergo rapid reaction, exchanging ligands very readily. Such complexes are called labile. With Cu(II) ion and NH3 in solution, one can, depending on the NH3 concentration, have one, two, three, or four NH3 molecules in the complex, in an equilibrium mixture. Most, but by no means all, complexes are labile. Some, including the one you will be making, exchange ligands only very slowly. For such species, the preparation reaction takes time, but once formed, the complex ion in solution may resist change rather dramatically. Such complexes are called inert. In the complex ion you will be making, NH3 and Cl− may be ligands. Since the overall charge of the compound must be zero, the number of moles of Cl− ion in a mole of compound will be determined by the charge on the cobalt ion. Some of the Cl− ions may be in the complex and some may simply be anions in the crystal. When the compound is dissolved in water, the free anions will go into solution, but the ones in the complex will remain firmly attached to cobalt. They will not react with a precipitating agent such as Ag+ ion, which will tend to form AgCl with any free Cl− ions. All of the NH3 molecules will be in the complex ion, and will not react with added H+ ion, as they would if they were free. To ensure that we find all the Cl− and NH3 that is present in the compound, we will destroy the complex before attempting any analyses. The complete analysis involves a gravimetric procedure for chloride ion, a colorimetric method for cobalt ion, and a volumetric procedure for ammonia. In some of the earlier experiments we have used these techniques (Exps. 4, 7, 23, 24, and 28), so you may be somewhat familiar with them. From the analyses we can determine the number of moles of each species present in a mole of compound. Because the mole ratio must equal the atom or molecule ratio, we can determine the formula of the compound. Although we describe the procedures for analysis of all three species in the compound, the formula can be determined from only two analyses, since the amount of the third species can be found by difference. The synthesis of the coordination compound will take one lab period. The analysis for both chloride and cobalt content can be done in one period. Determining the amount of ammonia present in the complex will take most of one period.

S


232

Experiment 29

Synthesis and Analysis of a Coordination Compound

Experimental Procedure Carry out the procedure in either A.1 or A.2, as directed by your instructor.

A.1 Preparation of Cox(NH3)yClz (Procedure I) Using a piece of weighing paper and a top-loading balance, weigh out 10 ± 0.2 g of ammonium chloride, NH4Cl. Pour the NH4Cl into a 250-mL beaker and add 40 mL distilled water. To this add 8 ± 0.2 g of cobalt(II) chloride hexahydrate, CoCl2⋅ H2O⋅ and 0.8 g activated charcoal (Norite). Heat this mixture to the boiling point, stirring to dissolve the soluble components. Cool the beaker in water, then in an ice bath. Slowly, in a hood, add 40 mL 15 M NH3, ammonia. CAUTION: This is a concentrated basic reagent, with a strong pungent odor that can knock you unconscious. After stirring, add 50 mL 10% H2O2, hydrogen peroxide, slowly, a few mL at a time, while continuing to stir. C A U T I O N : Be very careful not to spill this reagent on your skin. Use gloves. When the bubbling has stopped, place the 250-mL beaker in a 600-mL beaker containing 100 mL of water at about 60°C. Leave the beaker in the water bath for 30 to 40 minutes, holding the temperature of the bath at about 60°C as long as the liquid has a pink color. Stir occasionally. While the mixture is reacting, prepare a fritted glass filter crucible. Fit the crucible on a filter flask; clean the crucible by pulling, under suction, about 15 mL 1 M HNO3 through the disk, followed by 50 to 75 mL of distilled water. Put the crucible in an oven at 150°C for about an hour to dry. Prepare a second crucible the same way, and put it in the oven. When the 40 minutes are up, remove the 250-mL beaker from the water bath and cool it, first in a water bath and then in an ice bath, until the liquid in the beaker has a temperature below 5°C. Hold the mixture at this temperature for at least 5 minutes, stirring occasionally to promote crystallization of the crude product. Set up a Büchner funnel and, with suction, filter the mixture through the filter paper in the funnel. Discard the filtrate in the waste crock. Scrape the solid from the filter paper into the 250-mL beaker. Add 100 mL distilled water and, in a hood, 4 mL 12 M HCl. C A U T I O N : This reagent is a concentrated acid with a choking odor. Heat the mixture to boiling, with stirring, to dissolve the crystals. Rinse out the suction flask with distilled water, and reassemble the Büchner funnel. Holding the beaker with a pair of tongs or a folded paper towel, filter the hot mixture through the paper in the funnel, under suction. In this operation, the charcoal is removed and should be on the filter paper. The filtrate should be golden yellow and contains the product we seek. Transfer the filtrate to the (rinsed-out) 250-mL beaker and, in the hood, add 15 mL 12 M HCl. Place the beaker in an ice bath and stir for 5 minutes or so to promote formation of the golden crystals of the product, Cox(NH3)yClz. Check to see that the temperature is below 5°C. Pour about 50 mL of ice-cold distilled water on to a piece of filter paper in the Büchner funnel. After 2 or 3 minutes, turn on the suction and pull the water into the suction flask. Empty the flask. Then filter the cold mixture containing the product through the funnel, using suction. Turn the suction off, and pour about 20 mL 95% ethanol onto the crystals. Wait about 10 seconds, and then turn on the suction to pull through the ethanol, which should carry with it most of the water and HCl remaining on the crystals. Draw air through the crystals for several minutes. Weigh a 100-mL beaker to 0.1 g. Transfer the crystals from the filter paper to the beaker. Put the beaker in your locker. Take the fritted glass crucibles out of the oven, using tongs or a folded paper towel, and put the crucibles on a paper towel in your locker. Discard the liquid in the suction flask in the waste crock or as directed by your instructor.

A.2 Alternative Procedure for Preparation of Cox(NH3)yClz (Procedure II) There are several possible Cox(NH3)yClz compounds. The following procedure will allow you to make a different one from that produced in Part A.1 of this experiment.


Experiment 29


233

Using a piece of weighing paper and a top-loading balance, weigh out 4.0 ± 0.1 g of ammonium chloride, NH4Cl. Transfer the NH4Cl into a 250-mL beaker, and in the hood, slowly, add 25 mL of 15 M NH3. C A U T I O N : This is a concentrated basic reagent, with a strong pungent odor that can knock you unconscious. Cover the beaker with a watch glass. Add to the solution 8.0 ± 0.2 g of CoCl2.6H2O, cobalt(II) chloride hexahydrate, while stirring the mixture with a stirring rod. Continue stirring until most of the solid has dissolved and a slurry of fine tan solid in a deep red solution results. Measure out 20 mL of 10% hydrogen peroxide, H2O2. C A U T I O N : This reagent can cause severe skin burns; use gloves when handling it. In the hood, add the H2O2 to the mixture in 1-2 mL portions while stirring. There will be some effervescence (bubbling) as oxygen gas is evolved. Stir until bubbles of oxygen decrease significantly and all of the solid material dissolves. The initial Co2+ has now been oxidized to Co3+ and is present in the form of a complex ion. In the hood, measure out 25 mL of 12 M HCl. C A U T I O N : This is a concentrated acidic reagent with a choking odor. Add the HCl in approximately 5 mL increments while stirring the mixture. The white cloud that forms is solid ammonium chloride, formed by a gaseous acidbase neutralization reaction, dispersed in the air. Preferably in the hood, or at your lab bench, heat the mixture in the beaker to 75 to 85ºC for about 30 minutes, stirring occasionally. Keep the beaker covered with your watch glass. Control the temperature by judicious use of your Bunsen burner. A deep purple solid should gradually form in the beaker. This is your product, which may need to be recrystallized. While the mixture is being heated, prepare two fritted glass crucibles, according to the directions in the 4th paragraph of Part A.1 of this experiment. Remove heat from the beaker after 30 minutes and allow it to cool in a cool water bath—a 400 mL beaker containing 150 mL of cold tap water—for two minutes. Set up a vacuum filter apparatus, using a 2-part Büchner funnel. Transfer the smaller beaker to an ice bath. Keep the reaction beaker in the bath until the solution is colder than 10ºC. Using #1 filter paper that has been moistened while in the Büchner funnel, filter the solution, as demonstrated by your instructor. It is not necessary to transfer all of the solid. Knock or scrape the contents of the filter cup, including the paper, back into the 250 mL beaker. Rinse the cup with a few mL of water from your wash bottle. Similarly, using tweezers, rinse any remaining solid from the filter paper into the beaker, and dispose of the paper as directed by your instructor. The filtrate solution should be transferred to a designated waste container. It may be desirable to recrystallize your product to ensure its purity. If so advised by your lab instructor, and time is available, proceed as follows: Add water to the beaker from your wash bottle to bring the volume up to 125 mL, using the markings on the beaker. Stir the solution and take the beaker to the hood. In the hood, add 5 mL of 15 M NH3 to the beaker, in small portions, while stirring. Heat gently, if necessary, to dissolve all of the solid. Add 15 mL of 12 M HCl, in small portions, to the reaction beaker, while stirring. Preferably in the hood, or at your lab bench, heat the solution for 30 minutes as before but now at approximately 60ºC, stirring occasionally and using a watch glass to cover the beaker. At the end of the heating period, place the beaker in a cool water bath (as above) for two minutes. Cool further in an ice bath with occasional stirring, until the temperature falls below 10ºC, and for an additional five minutes. You have now recrystallized the desired cobalt-containing product, which should be a violet color. Reassemble your Büchner filter apparatus. To transfer as much of the product as possible, stir the mixture with your stirring rod just before quickly pouring out small portions of the suspension. Use a minimal amount of water from your wash bottle to transfer the remaining solid. Wash the product on the funnel with two five mL portions of water to aid in removing ammonium chloride. Wash with 10 mL of ethanol. Draw air through your product for five minutes, while weighing a watch glass to ± 0.01 g on a top-loading balance. Transfer the crystals to the watch glass and scrape any residual product from the filter paper onto the watch glass. Weigh the watch glass again and calculate your yield. Place the watch glass in your drawer until next lab period, at which time you will be analyzing your product. Take the fritted glass crucibles out of the oven, using tongs or a folded paper towel. Put the crucibles, after cooling, in your drawer on a paper towel. DISPOSAL OF REACTION PRODUCTS. Discard the liquid in the suction flask into the waste container or as otherwise directed by your instructor. Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

234

Experiment 29


B. Gravimetric Determination of Chloride Content Weigh your prepared compound in the 100-mL beaker to 0.0001 g on an analytical balance. (If a week has gone by, the compound should be well dried. If you are attempting to proceed on the same day as you prepared the compound, you must dry it thoroughly. This could be done by putting the sample under vacuum, while keeping the sample warm. If the sample does not lose 0.0001 g in 2 minutes while on the analytical balance, you may presume that it is dry.) Transfer 0.3 ± 0.05 g of the compound from the beaker to a labeled 250-mL beaker, and reweigh the 100-mL beaker to 0.0001 g. Transfer a second sample of about the same size to another labeled 250-mL beaker, and again weigh the 100-mL beaker and the remaining compound accurately. Record all masses on the Data page. Cover the 100-mL beaker with a watch glass and put it in your locker. Add 25 mL of distilled water to the sample in one of the 250-mL beakers. Add 5 mL 6 M NaOH and stir the mixture until all of the solid has dissolved. Heat the beaker and its contents to boiling, and simmer for 3 minutes, stirring occasionally. This step will destroy the complex; the mixture will turn black because of the formation of Co3O4. Once the black Co3O4 is formed, turn off the heat and let the beaker cool for a few minutes. Then add 6 M HNO3 until the mixture is acidic to litmus (it will take about 5 mL). Add 1 mL more of the HNO3. Add a small scoop of solid Na2SO3, sodium sulfite (approximately 0.2 g) and stir; this will reduce the Co3O4 and produce the Co2+ ion. The solution should clarify and turn pink. Wash down the walls of the beaker with water to bring any unreacted dark material into contact with the solution. Reheat to boiling and simmer gently for a minute or so. Then add 75 mL of distilled water. Slowly, with stirring, add 50 mL 0.1 M AgNO3, silver nitrate. This will precipitate all of the chloride ion in the solution as silver chloride, AgCl. Heat the mixture, with occasional stirring, to the boiling point. This will help coagulate the AgCl crystals and so facilitate filtration. Remove the beaker from the wire gauze and cover with a watch glass. Let it cool for 10 minutes or so. While the beaker is cooling, take one of the fritted glass crucibles you prepared in the previous session from your locker and weigh it to 0.0001 g on the analytical balance. Then set up the filter crucible apparatus. Carefully, with suction on, transfer all of the AgCl precipitate to the filter crucible. Use your wash bottle and rubber policeman to complete the transfer. Wash the AgCl, with suction on, with 10 mL H2O, followed by 5 mL 6 M HNO3, followed by 20 mL H2O. Keep the suction on for a minute after the last liquid has been pulled through the filter. Put the crucible and its contents in an oven at 150°C for at least an hour. Then let it cool to room temperature in a covered beaker. When it is cool, weigh it to 0.0001 g on the analytical balance. Put the solid AgCl in the AgCl crock. Repeat the entire procedure, using the second sample of compound.

C. Colorimetric Determination of Cobalt Content Transfer 0.5 ± 0.05 g of your compound into a clean, dry 50-mL beaker, using the same procedure as in the chloride determination. Weighings should be made to 0.0001 g. Make another transfer of about 0.5 g into another 50-mL beaker, again weighing the 100-mL beaker and its contents accurately. Cover each beaker with a watch glass and heat one of them gently with the Bunsen burner until the solid melts, foams, and turns blue. In this process the complex is destroyed and cobalt ion freed from the ligands. Allow the beaker and its contents to cool. Repeat the procedure with the sample in the other beaker. To the solid in the first beaker add 5 mL 6 M H2SO4 and 5 mL distilled water. Heat the beaker until the liquid begins to boil and the solid is all dissolved. Wash any deposits on the watch glass into the beaker with a few milliliters of water from your wash bottle. Repeat these steps with the sample in the other beaker. Transfer the solution from the first beaker to a clean 25-mL volumetric flask. Rinse the beaker with small portions of distilled water, adding the washings to the volumetric flask, so that all of the solution goes into the flask. Fill the flask to the mark with distilled water from your wash bottle. Stopper the flask and invert it at least twenty times to ensure that the solution inside becomes thoroughly mixed. Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Experiment 29


235

Pour some of the solution in the flask into a spectrophotometric test tube (3/4 full). Measure and record the absorbance of the solution at 510 nm. From the absorbance and the calibration curve furnished by your instructor, determine the molarity of cobalt ion in the solution. DISPOSAL OF REACTION PRODUCTS.

Pour the rest of the solution in the volumetric flask in the

waste crock.

Rinse the flask, and use it with the solution you prepared in the other beaker. Treat that solution as you did the first one, recording its absorbance and molarity of cobalt ion on the Data page.

D. Volumetric Determination of Ammonia Content In this part of the experiment we will first decompose the complex, releasing NH3 to the solution, as we did in the chloride analysis. We then distill off the NH3 into another container, and titrate it against an acid. The procedure is called a Kjeldahl analysis, and is used for determining nitrogen content of many organic and inorganic substances. Assemble the distillation apparatus shown in Figure 29.1. The details of the apparatus will vary from school to school but you will need a 250-mL distilling flask, a distilling head, a condenser, and a receiver adapter, in addition to a 125-mL Erlenmeyer flask, which will serve as a receiving flask. The 250-mL flask should be on a wire gauze on an iron ring. The flask and condenser should be held with clamps, which are to be adjusted so that all joints are tight. On the end of the receiver adapter attach a 4-inch length of latex tubing, which should reach to the bottom of the receiving flask. Put 50 mL saturated boric acid solution in the 125-mL receiving flask; add five drops bromcresol green indicator, and adjust the level of the flask, if necessary, so that the latex tubing is well under the liquid surface.

Figure 29.1


236

Experiment 29


Weigh out 1 ± 0.1 g of your compound into a beaker, as you did earlier, making the weighing to 0.0001 g. Disconnect the distilling flask from the head. Transfer the sample to the distilling flask, using a funnel. Rinse the beaker several times with small portions of distilled water, and add the washings to the flask. All the sample must end up in the flask. Add distilled water as necessary to give a final volume of about 50 mL. Swirl the flask to dissolve the sample. Add a few boiling chips and several pieces of granulated zinc. Start water flowing through the condenser, slowly. If you are working with standard taper glassware, lightly grease the lower joint of the distilling head. Pour 40 mL 6 M NaOH and 50 mL distilled water into the flask and reconnect it promptly to the distilling head. Make sure that all joints are tight, and that the top of the distilling head is stoppered. Turn on the Bunsen burner and bring the liquid in the flask to a boil. The complex will be destroyed, the liquid will turn dark as Co3O4 forms, and NH3 will be driven from the solution into the receiving flask. Adjust the burner as necessary to maintain smooth boiling and minimize bumping. The color of the indicator will change as the NH3 is absorbed. There may be a tendency for liquid to rise in the latex tubing during the distillation; this is not serious, but if the level goes up more than a few inches, you can let in a little air by momentarily loosening one of the joints. When the volume of liquid in the receiving flask reaches 100 mL, 50 mL of the solution will have been distilled and essentially all of the NH3 driven into the boric acid. With the burner still heating the flask, disconnect the receiver adapter from the condenser. (If the heat is turned off first, liquid will tend to back up into the condenser.) Rinse the adapter with the tubing still in the flask, washing any liquid on the inner or outer surface into the distillate. Then turn the burner off. Pour the distillate into a 250-mL volumetric flask. Rinse the receiving flask with a few small portions of distilled water, adding the washings to the volumetric flask. Fill to the mark with distilled water. Stopper the flask and invert it at least 20 times to ensure that the liquid is thoroughly mixed. Clean two burets. Rinse one of them with several small portions of the NH3 solution in the volumetric flask, draining some of the solution through the stopcock. Fill the buret with the NH3 solution and read and record the level. Rinse the other buret with small portions of the standardized HCl solution from the stock supply, and then fill that buret with that solution. Record the level. Draw about 40 mL of the NH3 solution into a 250-mL Erlenmeyer flask, and add five drops of bromcresol green indicator. Titrate with HCl to the end point, where the indicator changes from blue to yellow. The actual end point is green and can be reached by back titrating as necessary with NH3. Record the final levels in the NH3 and HCl burets. Repeat the titration once or twice with 40-mL portions of the NH3 solution. The NH3-HCl volume ratios should check within 1% if all goes well. DISPOSAL OF REACTION PRODUCTS. The titrated solutions may be poured down the sink. The liquid remaining in the distilling flask should be discarded in the waste crock.


Name __________________________________

Section __________________________________

Experiment 29

Data and Calculations: Synthesis and Analysis of a Coordination Compound A.1 or A.2 Preparation of Cox(NH3)yClz Procedure used ____________ Mass of CoCl2 ⋅ 6 H2O

____________ g

Mass of 100-mL beaker

____________ g

Approximate mass of product (see Part B, line 1)

____________ g

B. Determination of Chloride Ion Mass of 100-mL beaker plus product

____________ g

Mass after first transfer

____________ g

Mass of sample I ____________ g

Mass after second transfer

____________ g

Mass of sample II ____________ g

Mass of first filter crucible

____________ g

Mass of second filter crucible

____________ g

Mass of first crucible plus AgCl

____________ g

Mass of AgCl I

____________ g

Mass of second crucible plus AgCl

____________ g

Mass of AgCl II

____________ g

C. Determination of Cobalt Ion Mass of 100-mL beaker plus product

____________ g

Mass after first transfer

____________ g

Mass of sample I ____________ g

Mass after second transfer

____________ g

Mass of sample II ____________ g

Absorbance of first solution

____________

Molarity of Co ion

____________ M

Absorbance of second solution

____________

Molarity of Co ion

____________ M

D. Determination of Ammonia Mass of 100-mL beaker plus product

____________ g

Mass after transfer

____________ g

Initial reading NH3 buret

____________ mL

____________ mL

Final reading NH3 buret

____________ mL

____________ mL

Initial reading HCl buret

____________ mL

____________ mL

Final reading HCl buret

____________ mL

____________ mL

Molarity of standardized HCl

____________ M

2nd Trial


238

Experiment 29


E. Calculation of Chloride Content Sample I

Sample II

Mass of AgCl

____________ g

____________ g

Mass of Cl− in AgCl

____________ g

____________ g

Mass of sample

____________ g

____________ g

Mass of Cl− per 100-g sample

____________ g

____________ g

Moles Cl− per 100-g sample

____________

____________

F. Calculation of Cobalt Content Sample I Molarity of cobalt ion

Sample II

____________ M

____________ M

____________

____________

Mass of sample

____________ g

____________ g

Moles cobalt per 100-g sample

____________

____________

Moles cobalt in 25 mL = moles cobalt in sample

G. Calculation of Ammonia Content Mass of sample ____________ g

Molarity of HCl Trial I

____________ M Trial II

Volume NH3 used

____________ mL

____________ mL

Volume HCl used

____________ mL

____________ mL

Moles HCl = moles NH3

____________

____________

= moles NH3 in sample

____________

____________

Moles NH3 per 100-g sample

____________

____________

Moles NH3 per 250-mL NH3 solution

H. Determination of the Formula of Cox(NH3)yClz For a 100-g sample:

Moles Cl− __________

Moles Co ion __________

Moles NH3 __________

Whole number ratio:

z = __________

x = __________

y = __________

Formula of compound:


Name __________________________________

Section __________________________________

Experiment 29

Advance Study Assignment: Synthesis and Analysis of a Coordination Compound A student prepared 7.2 g of Cox(NH3)yClz. She then analyzed the compound by the procedure in this experiment. A. In the gravimetric determination of chloride, she weighed out 0.2988 g of the compound. The following data were obtained: Mass of crucible plus AgCl

19.0020 g

Mass of crucible

18.4628 g

Mass of AgCl

______________ g

MM AgCl = 143.323 g

Mass of Cl− in AgCl

______________ g

MM Cl− = 35.453 g

Moles Cl− in sample

______________

Moles Cl− in 100-g sample

______________

B. In the colorimetric determination of cobalt, she used a sample weighing 0.5120 g. The molarity of cobalt ion in the solution from the volumetric flask was 0.085 M. Moles cobalt ion in 25 mL solution = moles cobalt in sample

______________

Moles cobalt per 100-g sample

______________

C. In the volumetric determination of ammonia, the sample weighed 0.9985 g. In the titration, 0.1000 M HCl was used. She found that 40.00 mL of the NH3 solution required 39.62 mL of the HCl to reach the end point. No. moles HCl used

______________ = no. moles NH3 in 40 mL NH3 solution

No. moles NH3 in 250 mL NH3

______________ = no. moles NH3 in sample

No. moles NH3 per 100-g sample

______________

D. Calculation of the formula of Cox(NH3)yClz: In 100 g of sample,

Dividing by smallest number,

no. moles Co ion = ______________ no. moles NH3 =

______________

no. moles Cl− =

______________

no. moles Co ion

______________

no. moles NH3

______________

no. moles Cl−

______________

Formula of compound ______________ (This may or may not be the formula of the compound you will be making!)



Experiment 30 Determination of Iron by Reaction with Permanganate—A Redox Titration

P

otassium permanganate, KMnO4, is widely used as an oxidizing agent in volumetric analysis. In acid solution, MnO4− ion undergoes reduction to Mn2+ as shown in the following equation: 8 H + (aq ) + MnO 4 − (aq ) + 5 e − → Mn 2+ (aq ) + 4 H 2O

Since the MnO4− ion is violet and the Mn2+ ion is nearly colorless, the end point in titrations using KMnO4 as the titrant can be taken as the first permanent pink color that appears in the solution. KMnO4 will be employed in this experiment to determine the percentage of iron in an unknown containing iron(II) ammonium sulfate, Fe(NH4)2(SO4)2 ⋅ 6 H2O. The titration, which involves the oxidation of Fe2+ ion to Fe3+ by permanganate ion, is carried out in sulfuric acid solution to prevent the air oxidation of Fe2+. The end point of the titration is sharpened markedly if phosphoric acid is present, because the Fe3+ ion produced in the titration forms an essentially colorless complex with the acid. The number of moles of potassium permanganate used in the titration is equal to the product of the molarity of the KMnO4 and the volume used. The number of moles of iron present in the sample is obtained from the balanced equation for the reaction and the amount of MnO4− ion reacted. The percentage by weight of iron in the solid sample follows directly. You will calculate the mean percentage of iron for your values, along with the standard deviation. (See Appendix VIII.)

Experimental Procedure Obtain from the stockroom a buret and an unknown iron(II) sample. Weigh out accurately on the analytical balance three samples of about 1.0 g of your unknown into clean 250-mL Erlenmeyer flasks. Clean your buret thoroughly. Draw about 100 mL of the standard KMnO4 solution from the carboy in the laboratory. Rinse the buret with a few milliliters of the KMnO4 three times. Drain and then fill the buret with the KMnO4 solution. Prepare 150 mL of 1 M H2SO4 by pouring 25 mL of 6 M H2SO4 into 125 mL of H2O, while stirring. Add 50 mL of this 1 M H2SO4 to one of the iron samples. The sample should dissolve completely. Without delay, titrate this iron solution with the KMnO4 solution. When a light-yellow color develops in the iron solution during the titration, add 3 mL of 85% H3PO4. C A U T I O N : Caustic reagent. Continue the titration until you obtain the first pink color that persists for 15 to 30 seconds. Repeat the titration with the other two samples. Optional The KMnO4 solution can be standardized by the method you used in this experiment. Fe(NH4)2(SO4)2 ⋅ 6 H2O is a primary standard with a molar mass equal to 392.2 g. For standardization, use 0.7 ± 0.05 g samples of the primary standard, weighed to 0.0001 g. Draw 100 mL of the stock KMnO4 solution and titrate the samples as you did the unknown. Optional Experiment: Determine the mass percent of iron in iron supplement tablets that contain iron(II), ferrous ion. DISPOSAL OF REACTION PRODUCTS.

Dispose of your titrated solutions as directed by your labo-

ratory supervisor. Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.


Name ___________________________________ Section ___________________________________

Experiment 30

Data and Calculations: Determination of Iron by Reaction with Permanganate—A Redox Titration Mass of sample tube plus contents

______________ g

Mass after removing Sample I

______________ g

Mass after removing Sample II

______________ g

Mass after removing Sample III

______________ g

Molarity of standard KMnO4 solution

______________ M

Sample

I

II

III


______________ mL

______________ mL

______________ mL

Final buret reading

______________ mL

______________ mL

______________ mL

Volume KMnO4 required

______________ mL

______________ mL

______________ mL

Moles KMnO4 required

______________

______________

______________

Moles Fe2+ in sample

______________

______________

______________

Mass of Fe in sample

______________ g

______________ g

______________ g

Mass of sample

______________ g

______________ g

______________ g

Percentage of Fe in sample

______________ %

______________ %

______________ %

Unknown no. ______________

Average % Fe ______________ %

Standardization of KMnO4 Sample

Standard deviation ______________ %

Optional

IV

V

VI

Mass of primary standard

______________ g

______________ g

______________ g

Moles Fe in standard

______________

______________

______________ (continued on following page)


244

Experiment 30

Determination of Iron by Reaction with Permanganate—A Redox Titration

Moles KMnO4 required

______________

______________

______________


______________ mL

______________ mL

______________ mL

Final buret reading

______________ mL

______________ mL

______________ mL

Volume KMnO4 used

______________ mL

______________ mL

______________ mL

Molarity KMnO4

______________ M

______________ M

______________ M

Average molarity

______________ M


Name ___________________________________ Section ___________________________________

Experiment 30

Advance Study Assignment: Determination of Iron by Reaction with Permanganate—A Redox Titration 1. Write the balanced net ionic equation for the reaction between MnO4− ion and Fe2+ ion in acid solution.

2. How many moles of Fe2+ ion can be oxidized by 1.3 × 10−2 moles MnO4− ion in the reaction in Question 1?

______________ moles 3. A solid sample containing some Fe2+ ion weighs 1.264 g. It requires 38.67 mL 0.02487 M KMnO4 to titrate the Fe2+ in the dissolved sample to a pink end point. a. How many moles of MnO4− ion are required?

______________ moles b. How many moles of Fe2+ are there in the sample?

______________ moles c. How many grams of iron are there in the sample?

______________ g d. What is the mass percent of Fe in the sample?

______________ % 4. What is the mass percent of Fe in iron(II) ammonium sulfate hexahydrate, Fe(NH4)2(SO4)2 ⋅ 6 H2O?

______________ %



Experiment 31 Determination of an Equivalent Mass by Electrolysis

n Experiment 24 we determined the molar mass of an acid by titration of a known mass of the acid with a standardized solution of NaOH. We noted that the acid could contain only one acidic hydrogen atom per molecule if we were to be able to find the molar mass. If there are two or three such acidic hydrogen atoms we can only determine the mass of acid that will furnish one mole of H+ ion. That is called the equivalent mass of the acid. Experimentally we find that the equivalent mass of an element can be related in a fundamental way to the chemical effects observed in that phenomenon known as electrolysis. As you know, some liquids, because they contain ions, will conduct an electric current. If the two terminals on a storage battery, or any other source of DC voltage, are connected through metal electrodes to a conducting liquid, an electric current will pass through the liquid and chemical reactions will occur at the two metal electrodes; in this process electrolysis is said to occur, and the liquid is said to be electrolyzed. At the electrode connected to the negative pole of the battery, a reduction reaction will invariably be observed. In this reaction electrons will usually be accepted by one of the species present in the liquid, which, in the experiment we shall be doing, will be an aqueous solution. The species reduced will ordinarily be a metallic cation or the H+ ion or possibly water itself; the reaction that is actually observed will be the one that occurs with the least expenditure of electrical energy, and will depend on the composition of the solution. In the electrolysis cell we shall study, the reduction reaction of interest will occur in a slightly acidic medium; hydrogen gas will be produced by the reduction of hydrogen ion:

I

2 H+(aq) + 2 e− → H2(g)

(1)

In this reduction reaction, which will occur at the negative pole, or cathode, of the cell, for every H+ ion reduced one electron will be required, and for every molecule of H2 formed, two electrons will be needed. Ordinarily in chemistry we deal not with individual ions or molecules but rather with moles of substances. In terms of moles, we can say that, by Equation 1, The reduction of one mole of H+ ion requires one mole of electrons The production of one mole of H2(g) requires two moles of electrons A mole of electrons is a fundamental amount of electricity in the same way that a mole of pure substance is a fundamental unit of matter, at least from a chemical point of view. A mole of electrons is called a faraday, after Michael Faraday, who discovered the basic laws of electrolysis. The amount of a species that will react with a mole of electrons, or one faraday, is equal to the equivalent mass of that species. Since one faraday will reduce one mole of H+ ion, we say that the equivalent mass of hydrogen is 1.008 grams, equal to the mass of one mole of H+ ion (or 1/2 mole of H2(g)). To form one mole of H2(g) one would have to pass two faradays through the electrolysis cell. In the electrolysis experiment we will perform we will measure the volume of hydrogen gas produced under known conditions of temperature and pressure. By using the Ideal Gas Law we will be able to calculate how many moles of H2 were formed, and hence how many faradays of electricity passed through the cell. At the positive pole of an electrolysis cell (the metal electrode that is connected to the + terminal of the battery), an oxidation reaction will occur, in which some species will give up electrons. This reaction, which


248

Experiment 31

Determination of an Equivalent Mass by Electrolysis

takes place at the anode in the cell, may involve again an ionic or neutral species in the solution or the metallic electrode itself. In the cell that you will be studying, the pertinent oxidation reaction will be that in which a metal under study will participate: M(s) → Mn+(aq) + ne−

(2)

During the course of the electrolysis the atoms in the metal electrode will be converted to metallic cations and will go into the solution. The mass of the metal electrode will decrease, depending on the amount of electricity passing through the cell and the nature of the metal. To oxidize one mole, or one molar mass, of the metal, it would take n faradays, where n is the charge on the cation that is formed. By definition, one faraday of electricity would cause one equivalent mass, EM, of metal to go into solution. The molar mass, MM, and the equivalent mass of the metal are related by the equation: MM = EM × n

(3)

In an electrolysis experiment, since n is not determined independently, it is not possible to find the molar mass of a metal. It is possible, however, to find equivalent masses of many metals, and that will be our main purpose. The general method we will use is implied by the discussion. We will oxidize a sample of an unknown metal at the positive pole of an electrolysis cell, weighing the metal before and after the electrolysis and so determining its loss in mass. We will use the same amount of electricity, the same number of electrons, to reduce hydrogen ion at the negative pole of the electrolysis cell. From the volume of H2 gas that is produced under known conditions we can calculate the number of moles of H2 formed, and hence the number of faradays that passed through the cell. The equivalent mass of the metal is then calculated as the amount of metal that would be oxidized if one faraday were used. In an optional part of the experiment, your instructor may tell you the nature of the metal you used. Using Equation 3, it will be possible to determine the charge on the metallic cations that were produced during electrolysis.

Experimental Procedure Obtain from the stockroom a buret and a sample of metal unknown. Lightly sand the metal to clean it. Rinse the metal with water and then in acetone. Let the acetone evaporate. When the sample is dry, weigh it on the analytical balance to 0.0001 g. Set up the electrolysis apparatus as indicated in Figure 31.1. There should be about 100 mL 0.5 M HC2H3O2 in 0.5 M Na2SO4 in the beaker with the gas buret. This will serve as the conducting solution. Immerse the end of the buret in the solution and attach a length of rubber tubing to its upper end. Open the stopcock on the buret and, with suction, carefully draw the acid up to the top of the graduations. Close the stopcock. Insert the bare coiled end of the heavy copper wire up into the end of the buret; all but the coil end of the wire should be covered with watertight insulation. Check the solution level after a few minutes to make sure the stopcock does not leak. Record the level. The metal unknown will serve as the anode in the electrolysis cell. Connect the metal to the + pole of the power source with an alligator clip and immerse the metal but not the clip in the conducting solution. The copper electrode will be the cathode in the cell. Connect that electrode to the − pole of the power source. Hydrogen gas should immediately begin to bubble from the copper cathode. Collect the gas until about 50 mL have been produced. At that point, stop the electrolysis by disconnecting the copper electrode from the power source. Record the level of the liquid in the buret. Measure and record the temperature and the barometric pressure in the laboratory. (In some cases a cloudiness may develop in the solution during the electrolysis. This is caused by the formation of a metal hydroxide, and will have no adverse effect on the experiment.) Raise the buret, and discard the conducting solution in the beaker in the waste crock. Rinse the beaker with water, and pour in 100 mL of fresh conducting solution. Repeat the electrolysis, generating about 50 mL of H2 and recording the initial and final liquid levels in the buret. Take the alligator clip off the metal anode


Experiment 31

Determination of an Equivalent Mass by Electrolysis

249

Figure 31.1

and wash the anode with 0.1 M HC2H3O2, acetic acid. Rub off any loose adhering coating with your fingers, and then rinse off your hands. Rinse the electrode in water and then in acetone. Let the acetone evaporate. Weigh the dry metal electrode to the nearest 0.0001 g. When you are finished with the experiment, return the metal electrode to the stockroom and discard the conducting solution in the waste crock unless directed otherwise by your instructor.




Name ____________________________________

Section ________________________________

Experiment 31

Data and Calculations: Determination of an Equivalent Mass by Electrolysis Mass of metal anode

____________ g

Mass of anode after electrolysis

____________ g


____________ mL

Buret reading after first electrolysis

____________ mL

Buret reading after refilling

____________ mL

Buret reading after second electrolysis

____________ mL

Barometric pressure

____________ mm Hg

Temperature t

____________ °C

Vapor pressure of H2O at t

____________ mm Hg

Total volume of H2 produced, V

____________ mL

Temperature T

____________ K

Pressure exerted by dry H2: P = Pbar − VPH O 2 (ignore any pressure effect due to liquid levels in buret)

____________ mm Hg

No. moles H2 produced, n (use Ideal Gas Law, PV = nRT)

____________ moles

No. of faradays passed (no. of moles of electrons)

____________

Loss in mass by anode

____________ g

no. g lost ⎛ ⎞ Equivalent mass of metal ⎝ EM = no. faradays passed ⎠

____________ g

Unknown metal number

____________

Optional

Nature of metal ____________ MM ____________ g

Charge n on cation ____________ (Eq. 3)



Name ____________________________________

Section ________________________________

Experiment 31

Advance Study Assignment: Determination of an Equivalent Mass by Electrolysis 1. In an electrolysis cell similar to the one employed in this experiment, a student observed that his unknown metal anode lost 0.208 g while a total volume of 96.30 mL of H2 was being produced. The temperature in the laboratory was 25°C and the barometric pressure was 748 mm Hg. At 25°C the vapor pressure of water is 23.8 mm Hg. To find the equivalent mass of his metal, he filled in the blanks below. Fill in the blanks as he did. PH2 = Pbar − VPH O = ____________ mm Hg = ____________ atm 2

VH2 = ____________ mL = ____________ L T = ____________ K nH2 = ____________ moles

nH2 =

PV RT

(where P = PH2 )

1 mole H2 requires passage of ____________ faradays No. of faradays passed = ____________ Loss of mass of metal anode = ____________ g No. grams of metal lost per faraday passed =

no. grams lost = ____________ g = EM no. faradays passed

The student was told that his metal anode was made of iron. MM Fe = ____________ g. The charge n on the Fe ion is therefore ____________. (See Eq. 3.) 2. In standard, IUPAC units, the faraday is equal to 96,480 coulombs. A coulomb is the amount of electricity passed when a current of one ampere flows for one second. Given the charge on an electron, 1.6022 × 10−19 coulombs, calculate a value for Avogadro’s number. ____________



Experiment 32 Voltaic Cell Measurements

any chemical reactions can be classified as oxidation-reduction reactions, because they involve the oxidation of one species and the reduction of another. Such reactions can conveniently be considered as the result of two half-reactions, one of oxidation and the other reduction. In the case of the oxidation-reduction reaction

M

Zn(s) + Pb2+ (aq) → Zn2+ (aq) + Pb(s) that would occur if a piece of metallic zinc were put into a solution of lead nitrate, the two reactions would be Zn(s) → Zn2+ (aq) + 2 e− oxidation 2 e− + Pb2+ (aq) → Pb(s) reduction The tendency for an oxidation-reduction reaction to occur can be measured if the two reactions are made to occur in separate regions connected by a barrier that is porous to ion movement. An apparatus, called a voltaic cell, in which this reaction might be carried out under this condition is shown in Figure 32.1. If we connect a voltmeter between the two electrodes, we will find that there is a voltage, or potential, between them. The magnitude of the potential is a direct measure of the driving force or thermodynamic tendency of the spontaneous oxidation-reduction reaction to occur. If we study several oxidation-reduction reactions, we find that the voltage of each associated voltaic cell can be considered to be the sum of a potential for the oxidation reaction and a potential for the reduction reaction. In the Zn,Zn2+ || Pb2+,Pb cell we have been discussing, for example, Ecell = EZn,Zn2 + oxidation reaction + EPb2 + ,Pb reduction reaction

(1)

By convention, the negative electrode in a voltaic cell is taken to be the one from which electrons are emitted (i.e., where oxidation occurs). The negative electrode is the one that is connected to the minus pole of the voltmeter when the voltage is measured.

Figure 32.1


256

Experiment 32

Voltaic Cell Measurements

Since any cell potential is the sum of two electrode potentials, it is not possible, by measuring cell potentials, to determine individual absolute electrode potentials. However, if a value of potential is arbitrarily assigned to one electrode reaction, other electrode potentials can be given definite values, based on the assigned value. The usual procedure is to assign a value of 0.0000 volts to the standard potential for the electrode reaction 2 H+ (aq) + 2 e− → H2(g);

EH0 + ,H

2

red

= 0.0000 V

For the Zn,Zn2+ || H+,H2 cell, the measured potential is 0.76 V, and the zinc electrode is negative. Zinc metal is therefore oxidized, and the cell reaction must be Zn(s) + 2 H+ (aq) → Zn2+ (aq) + H2(g);

0 Ecell = 0.76 V

Given this information, one can readily find the potential for the oxidation of Zn to Zn2+. 0 0 0 Ecell = EZn,Zn 2 + oxid + EH + ,H 0 0.76 V = EZn,Zn 2+ oxid + 0.00 V;

2

red

0 EZn,Zn 2+ oxid = +0.76 V

If the potential for a half-reaction is known, the potential for the reverse reaction can be obtained by changing the sign. For example: if

0 EZn,Zn then 2+ oxid = +0.76 V,

if

0 EPb 2+ ,Pb red = +Y V,

then

0 EZn 2+ ,Zn red = −0.76 V

0 EPb,Pb 2+ oxid = −Y V

In the first part of this experiment you will measure the voltages of several different cells. By arbitrarily assigning the potential of a particular half-reaction to be 0.00 V, you will be able to calculate the potentials corresponding to all of the various half-reactions that occurred in your cells. In our discussion so far we have not considered the possible effects of such system variables as temperature, potential at the liquid-liquid junction, size of metal electrodes, and concentrations of solute species. Although temperature and liquid junctions do have a definite effect on cell potentials, taking account of their influence involves rather complex thermodynamic concepts and is usually not of concern in an elementary course. The size of a metal electrode has no appreciable effect on electrode potential, although it does relate directly to the capacity of the cell to produce useful electrical energy. In this experiment we will operate the cells so that they deliver essentially no energy but exert their maximum potentials. The effect of solute ion concentrations is important and can be described relatively easily. For the cell reaction at 25°C: aA(s) + bB+ (aq) → cC(s) + dD2+ (aq) 0 − Ecell = Ecell

0.0592 [D2+ ]d log n [B+ ]b

(2)

0 where Ecell is a constant for a given reaction and is called the standard cell potential, and n is the number of electrons in either electrode reaction. By Equation 2 you can see that the measured cell potential, Ecell, will equal the standard cell potential if the molarities of D2+ and B+ are both unity, or, if d equals b, if the molarities are simply equal to each other. We will carry out experiments under such conditions that the cell potentials you observe will be very close to the standard potentials given in the tables in your chemistry text. Considering the Cu,Cu2+ || Ag+,Ag cell as a specific example, the observed cell reaction would be

Cu(s) + 2 Ag+ (aq) → Cu2+ (aq) + 2 Ag(s) For this cell, Equation 2 takes the form 0 − Ecell = Ecell

0.0592 [Cu 2+ ] log 2 [Ag + ]2

(3)


Experiment 32


257

In the equation n is 2 because in the cell reaction two electrons are transferred in each of the two halfreactions. E 0 would be the cell potential when the copper and silver salt solutions are both 1 M, since then the logarithm term is equal to zero. If we decrease the Cu2+ concentration, keeping that of Ag+ at 1 M, the potential of the cell will go up by about 0.03 volts for every factor of ten by which we decrease [Cu2+]. Ordinarily it is not convenient to change concentrations of an ion by several orders of magnitude, so in general, concentration effects in cells are relatively small. However, if we should add a complexing or precipitating species to the copper salt solution, the value of [Cu2+] would drop drastically, and the voltage change would be appreciable. In the experiment we will illustrate this effect by using NH3 to complex the Cu2+. Using Equation 3, we can actually calculate [Cu2+] in the solution of its complex ion. It is very low. In an analogous experiment we will determine the solubility product of AgCl. In this case we will surround the Ag electrode in a Cu,Cu2+ || Ag+,Ag cell, with a solution of known Cl− ion concentration that is saturated with AgCl. From the measured cell potential, we can use Equation 3 to calculate the very small value of [Ag+] in the chloride-containing solution. In the two experiments in which we change the cation concentrations, first Cu2+ and then Ag+, we end up with systems in equilibrium. In the first case Cu2+ ion is in equilibrium with Cu(NH3)42+ and NH3. Using the cell potential, we can calculate [Cu2+]. From the way we made up the system we can calculate [Cu(NH3)42+] and [NH3]. If we wish to do so, we can use these data to find the dissociation constant for the Cu(NH3)42+ ion. Many equilibrium constants are found by experiments of this sort. In the last experiment we have an equilibrium system containing Ag+, Cl−, and AgCl(s). Here we can use the cell potential to find [Ag+]. There isn’t much Ag+ present, but there is a tiny bit, and the cell potential lets us find it. We know [Cl−] from the way we made up the mixture in the crucible. From these data we can easily calculate Ksp for AgCl: + − AgCl(s) → ← Ag (aq) + Cl (aq)

Ksp = [Ag+][Cl−]

(4)

Experimental Procedure You may work in pairs in this experiment.

A. Cell Potentials In this experiment you will be working with these seven electrode systems: Ag+,Ag(s) Cu2+,Cu(s) Fe3+,Fe2+,Pt Zn2+,Zn(s)

Br2(l),Br−,Pt Cl2(g, 1 atm),Cl−,Pt I2(s),I−,Pt

Your purpose will be to measure enough voltaic cell potentials to allow you to determine the electrode potentials of each electrode by comparing it with an arbitrarily chosen electrode potential. Using the apparatus shown in Figure 32.1, set up a voltaic cell involving any two of the electrodes in the list. About 10 mL of each solution should be enough for making the cell. The solute ion concentrations may be assumed to be one molar and all other species may be assumed to be at unit activity, so that the potentials of the cells you set up will be essentially the standard potentials. Measure the cell potential and record it along with which electrode has negative polarity. Pour both solutions into a beaker and rinse the crucible and 50-mL beaker with distilled water. In a similar manner set up and measure the cell voltage and polarities of other cells, sufficient in number to include all of the electrode systems on the list at least once. Do not combine the silver electrode system with any of the halogen electrode systems because a precipitate will form; for the same reason, do not use the copper and iodine electrodes in the same cell. Any other combinations may be used. The data from this part of the experiment should be entered in the first three columns of the table in Part A.1 of your report.


258

Experiment 32


B. Effect of Concentration on Cell Potentials Set up the Cu,Cu2+ || Ag+,Ag cell, using 10 mL of the CuSO4 solution in the crucible and 10 mL of AgNO3 in the beaker. Measure the potential of the cell. While the potential is being measured, add 10 mL of 6 M NH3 to the CuSO4 solution, stirring carefully with your stirring rod. Measure the potential when it becomes steady. 1. Complex Ion Formation.

Remove the crucible from the cell you have just studied and discard the Cu(NH3)42+. Clean the crucible by drawing a little 6 M NH3 through it, using the adapter and suction flask. Then draw some distilled water through it. Reassemble the Cu-Ag cell, this time using the beaker for the Cu-CuSO4 electrode system. Put 10 mL 1 M KCl in the crucible; immerse the Ag electrode in that solution. Add a drop of AgNO3 solution to form a little AgCl, so that an equilibrium between Ag+ and Cl− can be established. Measure the potential of this cell, noting which electrode is negative. In this case [Ag+] will be very low, which will decrease the potential of the cell to such an extent that its polarity may change from that observed previously. 2. Determination of the Solubility Product of AgCl.

DISPOSAL OF REACTION PRODUCTS. As you finish each voltage measurement, pour the two solutions into a beaker. At the end of the experiment, pour the contents of the beaker into the waste crock, unless directed otherwise by your instructor.


________________

________________

________________

________________

________________

________________

________________

1. ________________

2. ________________

3. ________________

4. ________________

5. ________________

6. ________________

7. ________________

________________

________________

________________

________________

________________

________________

________________

Negative Electrode

________________

________________

________________

________________

________________

________________

________________

Oxidation Reaction

________________

________________

________________

________________

________________

________________

________________

0 Eoxidation in volts

________________

________________

________________

________________

________________

________________

________________

Reduction Reaction

________________

________________

________________

________________

________________

________________

________________

0 Ereduction in volts


A. Noting that oxidation occurs at the negative pole in a cell, write the oxidation reaction in each of the cells. The other electrode system must undergo reduction; write the reduction reaction that occurs in each cell. 0 B. Assume that EAg + ,Ag = 0.00 volts (whether in reduction or oxidation). Enter that value in the table for all of the silver electrode systems you used 0 0 0 = Eoxidation + Ereduction in your cells. Since Ecell , you can calculate E 0 values for all the electrode systems in which the Ag,Ag+ system was involved. Enter those values in the table. 0 0 = − Ereduction C. Using the values and relations in Part B and taking advantage of the fact that for any given electrode system, Eoxidation , complete the 0 0 table of E values. The best way to do this is to use one of the E values you found in Part B in another cell with that electrode system. That 0 potential, along with Ecell , will allow you to find the potential of the other electrode. Continue this process with other cells until all the electrode potentials have been determined.

Calculations

Cell Potential, 0 Ecell (volts)

Electrode Systems Used in Cell

A.1. Cell Potentials

Name ____________________________________ Section ________________________________

Experiment 32

Data and Calculations: Voltaic Cell Measurements



260

Experiment 32


A.2. Table of Electrode Potentials 0 0 In Table A.1, you should have a value for Ered or Eoxid for each of the electrode systems you have studied. 0 0 0 Remembering that for any electrode system, Ered = − Eoxid , you can find the value for Ered for each system. List those potentials in the left column of the table below in order of decreasing value.

0 Ereduction 0 (EAg+ ,Ag = 0.00

Electrode Reaction in Reduction

V)

0 Ereduction 0 (EH+ ,H2 = 0.00

V)

________________________

________________________

________________________

________________________

________________________

________________________

________________________

________________________

________________________

________________________

________________________

________________________

________________________

________________________

________________________

________________________

________________________

________________________

________________________

________________________

________________________

0 The electrode potentials you have determined are based on EAg + ,Ag = 0.00 V. The usual assumption is that 0 = 0.80 V. Convert from one base to the other by adding EH0 + ,H = 0.00 volts, under which conditions EAg + ,Ag red 2 0.80 volts to each of the electrode potentials, and enter these values in the third column of the table. 0 Why are the values of Ered on the two bases related to each other in such a simple way?

B. Effect of Concentration on Cell Potentials 1. Complex ion formation: 0 , before addition of 6 M NH3 Potential, Ecell

____________V

Potential, Ecell, after Cu(NH3)42+ formed

____________V

Given Equation 3 0 − Ecell = Ecell

0.0592 [Cu 2+ ] log 2 [Ag + ]2

(3)

calculate the residual concentration of free Cu2+ ion in equilibrium with Cu(NH3)42+ in the solution in the crucible. Take [Ag+] to be 1 M. [Cu2+] = ____________ M (continued on following page) Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Experiment 32


261

2. Solubility product of AgCl: 0 Potential, Ecell , of the Cu, Cu2+ || Ag+, Ag cell (from B.1)

Potential, Ecell, with 1 M KCl present

____________V

Negative electrode____________

____________V

Negative electrode____________

Using Equation 3, calculate [Ag+] in the cell, where it is in equilibrium with 1 M Cl− ion. (Ecell in Equation 3 is the negative of the measured value if the polarity is not the same as in the standard cell.) Take [Cu2+] to be 1 M.

[Ag+] = ____________ M Since Ag+ and Cl− in the crucible are in equilibrium with AgCl, we can find Ksp for AgCl from the concentration of Ag+ and Cl−, which we now know. Formulate the expression for Ksp for AgCl, and determine its value. Ksp = ____________



Name ____________________________________

Section ________________________________

Experiment 32

Advance Study Assignment: Voltaic Cell Measurements 1. A student measures the potential of a cell made up with l M CuSO4 in one solution and l M AgNO3 in the other. There is a Cu electrode in the CuSO4 and an Ag electrode in the AgNO3, and the cell is set up 0 as in Figure 32.1. She finds that the potential, or voltage, of the cell, Ecell , is 0.45 V, and that the Cu electrode is negative. a. At which electrode is oxidation occurring?

____________

b. Write the equation for the oxidation reaction.

c. Write the equation for the reduction reaction.

0 d. If the potential of the silver, silver ion electrode, EAg + ,Ag is taken to be 0.000 V in oxidation or reduc0 0 0 0 tion, what is the value of the potential for the oxidation reaction, ECu ? Ecell . = Eoxid + Ered ,Cu 2 + oxid

____________ volts 0 e. If EAg + ,Ag red equals 0.80 V, as in standard tables of electrode potentials, what is the value of the poten0 tial of the oxidation reaction of copper, ECu ? ,Cu 2 + oxid

____________ volts f.

Write the net ionic equation for the spontaneous reaction that occurs in the cell that the student studied.

g. The student adds 6 M NH3 to the CuSO4 solution until the Cu2+ ion is essentially all converted to Cu(NH3)42+ ion. The voltage of the cell, Ecell, goes up to 0.92 V and the Cu electrode is still negative. Find the residual concentration of Cu2+ ion in the cell. (Use Eq. 3.)

____________ M h. In Part g, [Cu(NH3)42+] is about 0.05 M, and [NH3] is about 3 M. Given those values and the result in Part 1g for [Cu2+], calculate K for the reaction: 2+ Cu(NH3)42+(aq) → ← Cu (aq) + 4 NH3(aq)

____________



Experiment 33 Preparation of Copper(I) Chloride

xidation-reduction reactions are, like precipitation reactions, often used in the preparation of inorganic substances. In this experiment we will employ a series of such reactions to prepare one of the less commonly encountered salts of copper, copper(I) chloride. Most copper compounds contain copper(II), but copper(I) is present in a few slightly soluble or complex copper salts. The process of synthesis of CuCl we will use begins by dissolving copper metal in nitric acid:

O

Cu(s) + 4 H+ (aq) + 2 NO3− (aq) → Cu2+ (aq) + 2 NO2(g) + 2 H2O

(1)

The solution obtained is treated with sodium carbonate in excess, which neutralizes the remaining acid with evolution of CO2 and precipitates Cu(II) as the carbonate: 2 H+ (aq) + CO32− (aq)

(H2CO3)(aq)

Cu2+ (aq) + CO32− (aq)

CO2(g) + H2O

CuCO3(s)

(2) (3)

The CuCO3 will be purified by filtration and washing and then dissolved in hydrochloric acid. Copper metal added to the highly acidic solution reduces the Cu(II) to Cu(I) and is itself oxidized to Cu(I) in a disproportionation reaction. In the presence of excess chloride, the copper will be present as a CuCl43− complex ion. Addition of this solution to water destroys the complex, and white CuCl precipitates. CuCO3(s) + 2 H+ (aq) + 4 Cl− (aq) → CuCl42− (aq) + CO2(g) + H2O

(4)

CuCl42− (aq) + Cu(s) + 4 Cl− (aq) → 2 CuCl43− (aq)

(5)

2O CuCl 43− (aq ) ⎯H⎯ ⎯ → CuCl(s) + 3 Cl− (aq)

(6)

Because CuCl is readily oxidized, due care must be taken to minimize its exposure to air during its preparation and while it is being dried. In an optional part of the experiment, we prove that the formula of the compound synthesized is CuCl.

Experimental Procedure Obtain a 1-gram sample of copper metal turnings, a Buchner funnel, and a filter flask from the stockroom. Weigh the copper metal on the top-loading balance to 0.1 g. Put the metal in a 150-mL beaker and under a hood add 5 mL 15 M HNO3. C A U T I O N : Caustic reagent. Brown NO2 gas will be evolved and an acidic blue solution of Cu(NO3)2 produced. If it is necessary, you may warm the beaker gently with a Bunsen burner to dissolve all of the copper. When all of the copper is in solution, add 50 mL water to the solution and allow it to cool. Weigh out about 5 grams of sodium carbonate in a small beaker on the top-loading balance. Add small amounts of the Na2CO3 to the solution with your spatula, adding the solid as necessary when the evolution of CO2 subsides. Stir the solution to expose it to the solid. When the acid is neutralized, a blue-green precipitate of CuCO3 will begin to form. At that point, add the rest of the Na2CO3, stirring the mixture well to ensure complete precipitation of the copper carbonate. Transfer the precipitate to the Buchner funnel and use suction to remove the excess liquid. Use your rubber policeman and a spray from your wash bottle to make a complete transfer of the solid. Wash the


266

Experiment 33

Preparation of Copper(I) Chloride

precipitate well with distilled water with suction on; then let it remain on the filter paper with suction on for a minute or two. Remove the filter paper from the funnel and transfer the solid CuCO3 to the 150-mL beaker. Add 10 mL water and then 30 mL 6 M HCl slowly to the solid, stirring continuously. When the CuCO3 has all dissolved, add 1.5 g Cu turnings to the beaker and cover it with a watch glass. Heat the mixture in the beaker to the boiling point and keep it at that temperature, just simmering, for about 10 minutes. It may be that the dark-colored solution that forms will clear to a yellow color before that time is up, and if it does, you may stop heating and proceed to the next step. While the mixture is heating, put 150 mL distilled water in a 400-mL beaker and put the beaker in an ice bath. Cover the beaker with a watch glass. After you have heated the acidic Cu-CuCl2 mixture for 10 minutes or as soon as it turns light colored, carefully decant the hot liquid into the beaker of water, taking care not to transfer any of the excess Cu metal to the beaker. White crystals of CuCl should form. Continue to cool the beaker in the ice bath to promote crystallization and to increase the yield of solid. Cool 25 mL distilled water, to which you have added five drops 6 M HCl, in an ice bath. Put 20 mL acetone into a small beaker. Filter the crystals of CuCl in the Buchner funnel using suction. Swirl the beaker to aid in transferring the solid to the funnel. Just as the last of the liquid is being pulled through, wash the CuCl with 1/3 of the acidified cold water. Rinse the last of the CuCl into the funnel with another portion of the water and use the final 1/3 to rewash the solid. Turn off suction and add 1/2 of the acetone to the funnel; wait about 10 seconds and turn on the suction. Repeat this operation with the other half of the acetone. Draw air through the sample for a few minutes to dry it. If you have properly washed the solid, it will be pure white: if the moist compound is allowed to come into contact with air, it will tend to turn pale green, due to oxidation of Cu(I) to Cu(II). Weigh the CuCl in a previously weighed beaker to 0.1 g. Show your sample to your instructor for evaluation. DISPOSAL OF REACTION PRODUCTS. The contents of the suction flask after the first filtration (of CuCO3) can be discarded down the sink. The contents after the second filtration include an appreciable amount of copper, so they should be put in the waste crock.

Determination of the Formula of Copper(I) Chloride

Optional

There are many ways to prove that the formula of the copper compound you have prepared is indeed CuCl. The following procedure is easy to carry out and gives good results. Dry the copper compound by putting it under a heat lamp or in a drying oven at 110°C for 10 minutes. Weigh out about a 0.1 g sample of the compound accurately on the analytical balance, using a weighed 50-mL beaker as a container. Dissolve the sample in 5 mL of 6 M HNO3. Add 10 mL of distilled water, mix, and transfer the solution to a clean 100-mL volumetric flask. Use several 5-mL portions of distilled water to rinse the remaining solution from the beaker into the flask. Fill the flask to the mark with distilled water. Stopper the flask and invert it about 20 times to ensure that the solution in the flask is thoroughly mixed. Use a clean 10-mL pipet to transfer a 10-mL aliquot of the solution to a clean, dry 50-mL beaker. Add 10 mL 6 M NH3 to the beaker, using the 10-mL pipet, after you have rinsed it with 6 M NH3. This will convert any Cu2+ in the solution to deep blue Cu(NH3)42+. After mixing, measure the absorbance of the solution of Cu(NH3)42+ at a wavelength of 575 nm. Determine [Cu(NH3)42+] from a graph that is furnished to you or by comparison with a standard solution. (You can prepare the standard by accurately weighing about 0.06 g of copper turnings and putting them into a 50-mL beaker. Dissolve the copper in 5 mL of 6 M HNO3, and proceed as you did with the solution of the copper compound in the previous paragraph. Measure the absorbance of the standard at 575 nm. Calculate [Cu(NH3)42+] in the solution of the compound by assuming that Beer’s Law holds.) Knowing that concentration, and the fact that the original sample was effectively diluted to a volume of 200 mL, calculate the number of moles of Cu in the sample. Then calculate the number of grams of Cu in the sample, and the number of grams of Cl by difference. Use that value to find the number of moles of Cl, and from the mole ratio Cu:Cl obtain the formula of the compound.


Name ____________________________________

Section ________________________________

Experiment 33

Data and Calculations: Preparation of Copper(I) Chloride Mass of Cu sample

____________ g

Mass of beaker

____________ g

Mass of beaker plus CuCl

____________ g

Mass of CuCl prepared

____________ g

Theoretical yield

____________

Percentage yield

____________ %

Determination of the Formula of Copper(I) Chloride

Optional

Mass of Cu(I) chloride sample

____________ g

Absorbance of solution of Cu(NH3)42+

____________

Mass of Cu turnings

____________ g

Absorbance of standard solution

____________

No. moles Cu in turnings

____________ moles

All of the copper in the turnings is converted to Cu(NH3)42+ in a solution whose total volume would be 200 mL. [Cu(NH3)42+] in standard solution

____________ M

If Beer’s Law holds, [Cu(NH3)42+] in solution of sample = [Cu(NH3)42+] in standard × [Cu(NH3)42+] in solution of sample

Abs of sample Abs of standard ____________ M

No. moles Cu in sample (volume = 0.200 L)

____________ moles

No. grams Cu in sample

____________ g

No. grams Cl in sample (sample mass = mass Cu + mass Cl)

____________ g

No. moles Cl in sample

____________ g

Mole ratio Cl:Cu

____________

Formula of prepared compound (rounded to nearest integers)

____________



Name ____________________________________

Section ________________________________

Experiment 33

Advance Study Assignment: Preparation of Copper(I) Chloride 1. The Cu2+ ions in this experiment are produced by the reaction of 1.0 g of copper turnings with excess nitric acid. How many moles of Cu2+ are produced?

2. Why isn’t hydrochloric acid used in a direct reaction with copper to prepare the CuCl2 solution?

3. How many grams of metallic copper are required to react with the number of moles of Cu2+ calculated in Problem 1 to form the CuCl? The overall reaction can be taken to be: Cu2+ (aq) + 2 Cl− (aq) + Cu(s) → 2 CuCl(s) ____________ g 4. What is the maximum mass of CuCl that can be prepared from the reaction sequence of this experiment, using 1.0 g of Cu turnings to prepare the Cu2+ solution?

____________ g 5. A sample of the compound prepared in this experiment, weighing 0.1021 g, is dissolved in HNO3, and diluted to a volume of 100 mL. A 10-mL aliquot of that solution is mixed with 10 mL 6 M NH3. The [Cu(NH3)42+] in the resulting solution is found to be 5.16 × 10−3 M. a. How many moles of Cu were in the original sample, which had been effectively diluted to a volume of 200 mL? ____________ moles b. How many grams of Cu were in the sample? ____________ g c. How many grams of Cl were in the sample? How many moles? ____________ g ____________ moles d. What is the formula of the copper chloride compound?



Experiment 34 Development of a Scheme for Qualitative Analysis

n many of the previous experiments in this book you were asked to find out how much of a given species is present in a sample. You may have determined the concentration of chloride in an unknown solution, the molarity of an NaOH solution, and the amount of calcium ion in a sample of hard water. All of these experiments fall into that part of chemistry called quantitative analysis. Sometimes chemists are interested more in the nature of the species in a sample than the amount of those species. In that sort of problem we find out what the sample contains but not how much. For example, in Experiment 12 students are asked to determine which alkaline earth halide is present in a solution. That experiment is one involving qualitative analysis. This and the next four experiments deal with the qualitative analysis of solutions containing various anions and metallic cations. The procedures in qualitative analyses of this sort involve using precipitation reactions to remove the cations sequentially from a mixture. If the precipitate can contain only one cation under the conditions that prevail, that precipitate serves to prove the presence of that cation. If the precipitate may contain several cations, it can be dissolved and further resolved in a series of steps that may include acid-base, complex ion formation, redox, or other precipitation reactions. The ultimate result is a resolution of the sample into fractions that can contain only one cation, whose presence is established by formation of a characteristic precipitate or a colored complex ion. In this experiment you will be asked to develop a scheme for the qualitative analysis of four cations, using this systematic approach. The behavior of the cations toward a set of common test reagents differs from one cation to another and furnishes the basis for their separation. The cations we will study are Ba2+, Mg2+, Cd2+, and Al3+. The test reagents we will use are l M Na2SO4, l M Na2CO3, 6 M NaOH, 6 M NH3, and 6 M HNO3. These reagents furnish anions or molecules that will precipitate or form complex ions with the cations. So you may observe formation of insoluble sulfates, carbonates, and hydroxides (the last with either NaOH or NH3). In addition you may form complexes with OH− and NH3 when 6 M NaOH or 6 M NH3 is added to the cation-containing solutions. The complexes may be quite stable, stable enough to prevent precipitation of an otherwise insoluble salt on addition of a particular anion. The complexes, however, are all unstable in excess acid and can be broken down on addition of 6 M HNO3, releasing the cation for further reactions. In the first part of the experiment you will observe the behavior of the four cations in the presence of one or more of the reagents we have listed. On the basis of your observations you can set up the scheme for identifying the cations in a mixture. After testing the scheme with a known containing all of the cations, you will be given an unknown to analyze.

I

Experimental Procedure To four small test tubes add 1 mL of 0.1 M solutions of the nitrates or chlorides of Ba2+, Mg2+, Cd2+, and Al3+, one solution to a test tube (depth of 1 cm ≅ 1 mL). To each solution add one drop of 1 M Na2SO4, and stir with your stirring rod. Keep the rod in a 400-mL beaker of distilled water between tests. If a precipitate forms, write the formula of the precipitate in the box on the Data page corresponding to the cation-SO42− pair, to indicate that the sulfate of that cation is insoluble in water. Then add 1 mL more of the Na2SO4 to each test tube. If the precipitate dissolves on stirring, the cation forms a complex ion with sulfate ion. Under such conditions, put the formula of the complex ion in the box. In this experiment you can assume that any cations that form


272

Experiment 34

Development of a Scheme for Qualitative Analysis

complex ions have a coordination number of four. To any precipitates or complex ions, add 6 M HNO3, drop by drop, until the solution is acidic to litmus (blue to red). If the precipitate dissolves, note that with an A, meaning that the precipitate dissolves in acid. In the case of complexes, the precipitate that originally formed may reprecipitate if the ligands react with acid, and then dissolve again when the solution becomes acidic. If it reprecipitates, indicate that with a P, and as before use an A if the precipitate dissolves when the solution becomes acidic. Pour the contents of the four tubes into a beaker and rinse the tubes with distilled water. Repeat the tests, first with 1 M Na2CO3, then with 6 M NaOH, and finally with 6 M NH3. Although in most cases you will not observe formation of complex ions, you will see a few, and it is important not to miss them. One drop of reagent typically will produce a precipitate if the cation-anion compound is insoluble in water. The complex may be very stable and form readily in excess reagent. You may not get a precipitate reforming when you add HNO3. If the solution becomes acidic, and you saw no precipitate or a faint one, slowly add 1 M Na2CO3. If the carbonate comes down, write its formula in the box. When your table is complete you should be able to use it to state whether the sulfate, carbonate, and hydroxide of each of the cations is insoluble in water, and whether it dissolves in acid. You should also know whether the cation forms a complex ion with SO42−, CO32−, OH−, or NH3, and whether the complex is destroyed by acid. Now, given the solubility and reaction data you have obtained, your problem is to devise a step-by-step procedure for establishing whether each cation is present in a mixture. If you think about it for a while, several possible approaches should occur to you. As you construct your scheme, there are a few things to keep in mind besides the solubility and reaction data. 1. To separate a precipitate from a solution, you can use a centrifuge. Decant the solution into a test tube for use in further steps. See Appendix IV for some suggestions regarding procedures in qualitative analysis. 2. If a precipitate can contain only one cation, its presence serves to prove the presence of that cation. If it may involve more than one cation, it must be further resolved. In that case, the precipitate must be washed free of any cations that did not precipitate in that step because those cations would possibly interfere with later steps. To clean a precipitate, wash it twice with a 1:1 mixture of water and the precipitating reagent, stirring before centrifuging out the wash liquid. 3. pH is important. Your original tests were made with the cations in a water solution. If you want to obtain a precipitate, the solution must have a pH where that precipitate can form. 6 M HNO3 or 6 M NaOH can be used to bring a mixture to a pH of about 7. When you have your separation scheme in mind, describe it by constructing a flow diagram. The design of a flow diagram is discussed in the Advance Study Assignment. On the flow diagram the formulas of all species involving the cations should be given at the beginning and end of each step. Reagents are shown alongside the line connecting reactants and products. When you have completed your flow diagram, test your scheme with a known containing all four cations. If your scheme works, show your flow diagram to your instructor, who will give you an unknown to analyze. On completing the experiment, pour the contents of the beaker used for reaction products into the waste crock unless directed otherwise by your instructor.



Name ____________________________________

Section ________________________________

Experiment 34

Observations and Report Sheet: Development of a Scheme for Qualitative Analysis Table of Solubility Properties Ba2+

Mg2+

Cd2+

Al3+

1 M Na2SO4

1 M Na2CO3

6 M NaOH

6 M NH3

Key: No entry = soluble on mixing reagents Top formula = precipitate insoluble in water Second formula = complex ion that forms in excess reagent Bottom formula = carbonate that precipitates from acidified complex A = precipitate dissolves in acid P = precipitate reforms when complex is slowly acidified

Flow Diagram for Separation Scheme:

Observations on known: Observations on unknown: Cations present in unknown: ____________ ____________ ____________ ____________ Unknown no. ____________



Name ____________________________________

Section ________________________________

Experiment 34

Advance Study Assignment: Development of a Scheme for Qualitative Analysis Qualitative analysis schemes can be summarized by flow diagrams. The flow diagram for a scheme that might be used to analyze a mixture that could contain Cu2+, Pb2+, and Sn2+ is shown below:

The meaning of the flow diagram is almost obvious. In the first step, 6 M NaOH is added in excess. Cu(OH)2 precipitates, and Pb2+ and Sn2+ remain in solution because of the formation of hydroxo-complex ions. Cu(OH)2 is removed by centrifuging. In Step 2 the solution is made acidic with H2SO4, and PbSO4 precipitates after the complex is destroyed by the H+ ion in the acid. PbSO4 is removed by centrifuging. In Step 3 pH control is used to bring the pH to about 7. Then addition of Na2CO3 precipitates SnCO3.

1.

Construct the flow diagram for the separation scheme for a solution containing Ag+, Ni2+, and Zn2+. The steps in the procedure are as follows: Step 1. Add 6 M HCl to precipitate Ag+ as AgCl. Ni2+ and Zn2+ are not affected. Centrifuge out the

AgCl. Step 2. Add 6 M NaOH in excess, precipitating Ni(OH)2(s) and converting Zn2+ to the Zn(OH)42−

complex ion. Centrifuge out the Ni(OH)2. Step 3. Neutralize the solution with 6 M HCl to pH = 7. Add 1 M Na2CO3, precipitating ZnCO3.

Use the following page for your flow diagram.



Experiment 35 Spot Tests for Some Common Anions

here are two broad categories of problems in analytical chemistry. Quantitative analysis deals with the determination of the amounts of certain species present in a sample; there are several experiments in this manual involving quantitative analysis, and you probably have performed some of them. The other area of analysis, called qualitative analysis, has a more limited purpose, establishing whether given species are or are not present in detectable amounts in a sample. Several of the experiments in this manual have dealt with problems in qualitative analysis. One can carry out the qualitative analysis of a sample in various ways. Probably the simplest approach, which we will use in this experiment, is to test for the presence of each possible component by adding a reagent that will cause the component, if it is in the sample, to react in a characteristic way. This method involves a series of “spot” tests, one for each component, carried out on separate samples of the unknown. The difficulty with this way of doing qualitative analysis is that frequently, particularly in complex mixtures, one species may interfere with the analytical test for another. Although interferences are common, there are many ions that can, under optimum conditions at least, be identified in mixtures by simple spot tests. In this experiment we will use spot tests for the analysis of a mixture that may contain the following commonly encountered anions in solution:

T

CO32−

PO43−

Cl−

SCN−

SO42−

SO32−

C2H3O2−

NO3−

The procedures we will use involve simple acid-base, precipitation, complex ion formation, or oxidationreduction reactions. In each case you should try to recognize the kind of reaction that occurs, so that you can write the net ionic equation that describes it.

Experimental Procedure Carry out the test for each of the anions as directed. Repeat each test using a solution made by diluting the anion solution 9:1 with distilled water; use your 10-mL graduated cylinder to make the dilution and make sure you mix well before taking the sample for analysis. In some of the tests, a boiling-water bath containing about 100 mL water in a 150-mL beaker will be needed, so set that up before proceeding. When performing a test, if no reaction is immediately apparent, stir the mixture with your stirring rod to mix the reagents. These tests can easily be used to detect the anions at concentrations of 0.02 M or greater, but in dilute solutions careful observation may be required.

Test for the Presence of Carbonate Ion, CO 32Cautiously add 1 mL of 6 M HCl to 1 mL of 1 M Na2CO3 in a small test tube. With concentrated solutions, bubbles of carbon dioxide gas are immediately evolved. With dilute solutions, the effervescence will be much less obvious. Warming in the water bath, with stirring, will increase the amount of bubble formation. Carbon dioxide is colorless and odorless.


278

Experiment 35

Spot Tests for Some Common Anions

Test for the Presence of Sulfate Ion, SO 42Add 1 mL of 6 M HCl to 1 mL of 0.5 M Na2SO4. Add a few drops of 1 M BaCl2. A white, finely divided precipitate of BaSO4 indicates the presence of SO42− ion.

Test for the Presence of Phosphate Ion, PO 43Add 1 mL 6 M HNO3 to 1 mL of 0.5 M Na2HPO4. Then add 1 mL of 0.5 M (NH4)2MoO4 and stir thoroughly. A yellow precipitate of ammonium phosphomolybdate, (NH4)3PO4 ⋅ 12 MoO3, establishes the presence of phosphate. The precipitate may form slowly, particularly in the more dilute solution; if it does not appear promptly, put the test tube in the boiling-water bath for a few minutes.

Test for the Presence of Sulfite Ion, SO 32Sulfite ion in acid solution tends to evolve SO2, which can be detected by its odor even at low concentrations. Sulfite ion is slowly oxidized to sulfate in moist air, so sulfite-containing solutions will usually test positive for sulfate ion. To 1 mL 0.5 M Na2SO3 add 1 mL 6 M HCl, and mix with your stirring rod. Cautiously, sniff the rod to try to detect the acrid odor of SO2, which is a good test for sulfite. If the odor is too faint to detect, put the test tube in the hot-water bath for 10 seconds and sniff again. To proceed with a chemical test, add 1 mL 1 M BaCl2 to the solution in the tube. Stir, and centrifuge out any precipitate of BaSO4. Decant the clear solution into a test tube and add 1 mL 3% H2O2, hydrogen peroxide. Stir the solution and let it stand for a few seconds. If sulfite is present, its oxidation to sulfate will cause a new precipitate of BaSO4 to form. If thiocyanate is present, it may interfere with the chemical test. In that case, add 1 mL 1 M BaCl2 to 1 mL of the sample. Centrifuge out any precipitate, which will contain BaSO3 if sulfite is present, but will not contain thiocyanate. Wash the solid with 2 mL water, stir, centrifuge, and discard the wash. To the precipitate add 1 mL 6 M HCl and 2 mL water, and stir. Centrifuge out any BaSO4, and decant the clear liquid into a test tube. To the liquid add 1 mL 3% H2O2. If you have sulfite present, you will observe a new precipitate of BaSO4 within a few seconds.

Test for the Presence of Thiocyanate Ion, SCN− Add 1 mL 6 M acetic acid, HC2H3O2, to 0.5 M KSCN and stir. Add one or two drops 0.1 M Fe(NO3)3. A deep red coloration as a result of formation of FeSCN2+ ion is proof for the presence of SCN− ion.

Test for the Presence of Chloride Ion, Cl− Add 1 mL 6 M HNO3 to 1 mL 0.5 M NaCl. Add two or three drops 0.1 M AgNO3. A white, curdy precipitate of AgCl will form if chloride ion is present. Several anions interfere with this test, because they too form white precipitates with AgNO3 under these conditions. In this experiment only SCN− ion will interfere. If the sample contains SCN− ion, put 1 mL of the solution into a 30- or 50-mL beaker and add 1 mL 6 M HNO3. Boil the solution gently until its volume is decreased by half. This will destroy most of the thiocyanate. To the solution in a small test tube add 1 mL 6 M HNO3 and a few drops of AgNO3 solution. If Cl− is present you will get a curdy precipitate. If Cl− is absent, you may see some cloudiness due to residual amounts of SCN−.

Test for the Presence of Acetate Ion, C2H3O2To 1 mL 0.5 M NaC2H3O2 add 6 M NH3 or 6 HNO3 until the solution is just basic to litmus. Add one drop of 1 M BaCl2. If a precipitate forms, add 1 mL of the BaCl2 solution to precipitate interfering anions. Stir, centrifuge, and decant the clear liquid into a test tube. Add one drop BaCl2 to make sure that precipitation was complete.


Experiment 35


279

To 1 mL of the liquid add 0.1 M KI3, drop by drop, until the solution takes on a fairly strong rust color. Add 0.5 mL 0.1 M La(NO3)3 and six drops 6 M NH3. Stir, and put the test tube in the water bath. If acetate is present, the mixture will darken to nearly black in a few minutes. The color is due to iodine adsorbed on the basic lanthanum acetate precipitate.

Test for the Presence of Nitrate Ion, NO 3To 1 mL 0.5 M NaNO3 add 1 mL 6 M NaOH. Then add a few granules of Al metal, using your spatula, and put the test tube in the hot-water bath. In a few seconds, the Al-NaOH reaction will produce H2 gas, which will reduce the NO3− ion to NH3, which will come off as a gas. To detect the NH3, hold a piece of moistened red litmus paper just above the end of the test tube. If the sample contains nitrate ion, the litmus paper will gradually turn blue, within a minute or two. Blue spots caused by effervescence are not to be confused with the blue color over all of the litmus exposed to NH3 vapors. Cautiously sniff the vapors at the top of the tube; you may be able to detect the odor of ammonia. If SCN− is present, it will interfere with the test. In that case, first add 1 mL 1 M CuSO4 to 1 mL of the sample and put the test tube in the water bath for a minute or two. Centrifuge out the precipitate, and decant the solution into a test tube. Add 1 mL 1 Na2CO3 to remove excess Cu2+ ion. Centrifuge out the precipitate, and decant the solution into a test tube. To 1 mL of the solution add 1 mL 6 M NaOH, and proceed, starting with the second sentence of this procedure. When you have completed all of the tests, obtain an unknown from your laboratory supervisor, and analyze it by applying the tests to separate 1-mL portions. The unknown will contain three or four ions on the list, so your test for a given ion may be affected by the presence of others. When you think you have properly analyzed your unknown, you may, if you wish, make a “known” with the composition you found and test it to see if it behaves as your unknown did. DISPOSAL OF REACTION PRODUCTS. As you complete each test, pour the products into a beaker. When you have finished the experiment, pour the contents of the beaker into the waste crock, unless directed otherwise by your instructor.



Name ____________________________________

Section ________________________________

Experiment 35

Observations and Report Sheet: Spot Tests for Some Common Anions Observations and Comments on Spot Tests Ion

Stock Solution

9:1 Dilution

Unknown

CO32− ________________________________________________________________________________________ SO42− ________________________________________________________________________________________ PO43− ________________________________________________________________________________________ SO32− ________________________________________________________________________________________ Cl− ________________________________________________________________________________________ C2H3O2− ________________________________________________________________________________________ SCN− ________________________________________________________________________________________ NO3− ________________________________________________________________________________________

Unknown no. ____________ contains ________________________________________________________



Name ____________________________________

Section ________________________________

Experiment 35

Advance Study Assignment: Spot Tests for Some Common Anions 1. Each of the observations listed was made on a different solution. Given the observations, state which ion studied in this experiment is present. If the test is not definitive, indicate that with a question mark. A. Addition of 6 M NaOH and Al to the solution produces a vapor that turns red litmus blue. Ion present: B. Addition of 6 M HCl produces a vapor with an acrid odor. Ion present: C. Addition of 6 M HCl produces an effervescence. Ion present: D. Addition of 6 M HNO3 plus 0.1 M AgNO3 produces a precipitate. Ion present: E. Addition of 6 M HNO3 plus 1 M BaCl2 produces a precipitate. Ion present: F. Addition of 6 M HNO3 plus 0.5 M (NH4)2MoO4 produces a precipitate. Ion present: 2. An unknown containing one or more of the ions studied in this experiment has the following properties: A. No effect on addition of 6 M HNO3. B. No effect on addition of 0.1 M AgNO3 to solution in Part A. C. White precipitate on addition of 1 M BaCl2 to solution in Part A. D. Yellow precipitate on addition of (NH4)2MoO4, to solution in Part A. On the basis of this information which ions are present, which are absent, and which are in doubt? Present

Absent

In doubt


284

Experiment 35


3. The chemical reactions that are used in the anion spot tests in this experiment are for the most part simple precipitation or acid-base reactions. Given the information in each test procedure, try to write the net ionic equation for the key reaction in each test. A. CO32−

B. SO42−

C. PO43− (Reactants are HPO42−, NH4+, MoO42−, and H+; products are (NH4)3PO4 ⋅ 12 MoO3 and H2O; no oxidation or reduction occurs.)

D. SO32−

E. SCN−

F. Cl−

G. NO3− (Take as reactants Al and NO3−; as products NH3 and A1O2−; final equation also contains OH− and H2O as reactants.)


Experiment 36 Qualitative Analysis of Group I Cations

Precipitation and Separation of Group I Ions The chlorides of Pb2+, Hg22+, and Ag+ are all insoluble in cold water. They can be removed as a group from solution by the addition of HCl. The reactions that occur are simple precipitations and can be represented by the equations: Ag+ (aq) + Cl− (aq) → AgCl(s)

(1)

Pb2+ (aq) + 2 Cl− (aq) → PbCl2(s)

(2)

Hg22+ (aq) + 2 Cl− (aq) → Hg2Cl2(s)

(3)

It is important to add enough HCl to ensure complete precipitation, but not too large an excess. In concentrated HCl solution these chlorides tend to dissolve, producing chloro-complexes such as AgCl2−. Lead chloride is separated from the other two chlorides by heating with water. The PbCl2 dissolves in hot water by the reverse of Reaction 2: PbCl2(s) → Pb2+ (aq) + 2 Cl− (aq)

(4)

Once Pb2+ has been put into solution, we can check for its presence by adding a solution of K2CrO4. The chromate ion, CrO42−, gives a yellow precipitate with Pb2+: Pb2+ (aq) + CrO42− (aq) → PbCrO4(s) yellow

(5)

The other two insoluble chlorides, AgCl and Hg2Cl2, can be separated by adding aqueous ammonia. Silver chloride dissolves, forming the complex ion Ag(NH3)2+: AgCl(s) + 2 NH3 (aq) → Ag(NH3)2+ (aq) + Cl− (aq)

(6)

Ammonia also reacts with Hg2Cl2 via a rather unusual oxidation-reduction reaction. The products include finely divided metallic mercury, which is black, and a compound of formula HgNH2Cl, which is white: Hg2Cl2(s) + 2 NH3 (aq) → Hg(l) + HgNH2Cl(s) + NH4+ (aq) + Cl− (aq) white black white

(7)

As this reaction occurs, the solid appears to change color, from white to black or gray.


286

Experiment 36

Qualitative Analysis of Group I Cations

The solution containing Ag(NH3)2+ needs to be further tested to establish the presence of silver. The addition of a strong acid (HNO3) to the solution destroys the complex ion and reprecipitates silver chloride. We may consider that this reaction occurs in two steps: Ag(NH3)2+ (aq) + 2 H+ (aq) → Ag+ (aq) + 2 NH4+ (aq) Ag + (aq ) + Cl − (aq ) → AgCl(s) Ag( NH3 )2 (aq ) + 2 H + (aq ) + Cl − (aq ) → AgCl(s) + 2 NH 4 + (aq ) white +

(8)

Experimental Procedure See Appendix IV for some suggestions regarding procedures in qualitative analysis. Step 1. Precipitation of the Group I Cations. To gain familiarity with the procedures used in

qualitative analysis we will first analyze a known Group I solution, made by mixing equal volumes of 0.1 M AgNO3, 0.2 M Pb(NO3)2, and 0.1 M Hg2(NO3)2. Add two drops of 6 M HCl to 1 mL of the known solution in a small test tube. (1 mL ≅ 1 cm depth in the tube.) Mix with your stirring rod. Centrifuge the mixture, making sure there is a blank test tube containing about the same amount of water in the opposite opening in the centrifuge. Add one more drop of the 6 M HCl to test for completeness of precipitation. Centrifuge again if necessary. Decant the supernatant liquid into another test tube and save it for further tests if cations from other groups may be present. The precipitate will be white and will contain the chlorides of the Group I cations. Step 2. Separation of Pb2+. Wash the precipitate with 1 or 2 mL of water. Stir, centrifuge, and

decant the liquid, which may be discarded. Add 2 mL distilled water from your wash bottle to the precipitate in the test tube, and place the test tube in a 250-mL beaker that is about half full of boiling water. Leave the test tube in the bath for a minute or two, stirring occasionally with a glass rod. This will dissolve most of the PbCl2, but not the other two chlorides. Centrifuge the hot mixture, and decant the hot liquid into a test tube. Save the remaining precipitate for further tests. Step 3. Identification of Pb2+. Add one drop of 6 M acetic acid and a few drops of 0.1 M K2CrO4

to the solution from Step 2. If Pb2+ is present, a bright yellow precipitate of PbCrO4 will form. Step 4. Separation and Identification of Hg22+. To the precipitate from Step 2 add 1 mL 6 M NH3

and stir thoroughly. Centrifuge the mixture and decant the liquid into a test tube. A gray or black precipitate, produced by reaction of Hg2Cl2 with ammonia, proves the presence of Hg22+. Step 5. Identification of Ag+. Add 6 M HNO3 to the solution from Step 4 until it is acidic to litmus

paper. It will take about 1 mL. Test for acidity by dipping the end of your stirring rod in the solution, then touching it to a piece of blue litmus paper (red in acidic solution). If Ag+ is present, in the acidified solution it will precipitate as white AgCl. Step 6. When you have completed the tests on the known solution, obtain an unknown and analyze

it for the possible presence of Ag+, Pb2+, and Hg22+. DISPOSAL OF REACTION PRODUCTS.

All reaction products in this experiment should be dealt

with as directed by your instructor. Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Experiment 36

Qualitative Analysis of Group I Cations

287

Flow Diagrams It is possible to summarize the directions for analysis of the Group I cations in a flow diagram. In the diagram, successive steps in the procedure are linked with arrows. Reactant cations or reactant substances containing the ions are at one end of each arrow and products formed are at the other end. Reagents and conditions used to carry out each step are placed alongside the arrows. A partially completed flow diagram for the Group I ions follows:

You will find it useful to construct flow diagrams for each of the cation groups. You can use such diagrams in the laboratory as brief guides to procedure, and you can use them to record your observations on your known and unknown solutions.



Name ____________________________________

Section ________________________________

Experiment 36

Observations and Report Sheet: Qualitative Analysis of Group I Cations Flow Diagram for Group I

Observations on known (record on diagram if different from those on prepared diagram).

Observations on unknown (record on diagram in colored pencil to distinguish from observations on known).

Unknown no. ____________

Ions reported present ____________ ____________ ____________



Name ____________________________________

Section ________________________________

Experiment 36

Advance Study Assignment: Qualitative Analysis of Group I Cations 1. On the report sheet on page 289, complete the flow diagram for the separation and identification of the ions in Group I. 2. Write balanced net ionic equations for the following reactions: a. The precipitation of the chloride of Hg22+.

b. The dissolving of PbCl2 in hot water.

c. The dissolving of AgCl in aqueous ammonia.

3. You are given an unknown solution that contains only one of the Group I cations and no other metallic cations. Develop the simplest procedure you can think of to determine which cation is present. Draw a flow chart showing the procedure and the observations to be expected at each step with each of the possible cations. The information in Appendix IIA should be helpful.

4. A solution may contain Ag+, Pb2+, and Hg22+. A white precipitate forms on addition of 6 M HCl. The precipitate is insoluble in hot water. The residue turns black on addition of ammonia. Which of the ions are present, which are absent, and which remain undetermined? State your reasoning. Note: On paper unknowns such as this one, confirmatory tests are usually not included. Present ____________ Absent ____________ In doubt ____________



Experiment 37 Qualitative Analysis of Group II Cations

Precipitation and Separation of Group II Ions The sulfides of the four Group II ions, Bi3+, Sn4+, Sb3+, and Cu2+, are insoluble at a pH of 0.5. The solution is adjusted to this pH and then saturated with H2S, which precipitates Bi2S3, SnS2, Sb2S3, and CuS. The reaction with Bi3+ is typical: 2 Bi3+ (aq) + 3 H2S (aq) → Bi2S3(s) + 6 H+ (aq) black

(1)

Saturation with H2S could be achieved by simply bubbling the gas from a generator through the solution. A more convenient method, however, is to heat the acid solution after adding a small amount of thioacetamide. This compound, CH3CSNH2, hydrolyzes when heated in aqueous solution to liberate H2S: CH3CSNH2 (aq) + 2 H2O → H2S (aq) + CH3COO− (aq) + NH4+ (aq)

(2)

Using thioacetamide as the precipitating reagent has the advantage of minimizing odor problems and giving denser precipitates. The four insoluble sulfides can be separated into two subgroups by extracting with a solution of sodium hydroxide. The sulfides of tin and antimony dissolve, forming hydroxo-complexes: SnS2 (s) + 6 OH− (aq) → Sn(OH)62− (aq) + 2 S2− (aq)

(3)

Sb2S3 (s) + 8 OH− (aq) → 2 Sb(OH)4− (aq) + 3 S2− (aq)

(4)

Since Cu2+ and Bi3+ do not readily form hydroxo-complexes, CuS and Bi2S3 do not dissolve in solutions of NaOH. The solution containing the Sb(OH)4− and Sn(OH)62− complex ions is treated with HCl and thioacetamide. The H+ ions of the strong acid HCl destroy the hydroxo-complexes; the free cations then reprecipitate as the sulfides. The reaction with Sn(OH)62− may be written as: Sn(OH)62− (aq) + 6 H+ (aq) → Sn4+ (aq) + 6 H2O Sn 4+ (aq ) + 2 H 2S (aq ) → SnS2 (s) + 4 H + (aq ) Sn(OH)6 2− (aq ) + 2 H + (aq ) + 2 H 2S (aq ) → SnS2 (s) + 6 H 2O tan

(5)

The Sb(OH)4− ion behaves in a very similar manner, being converted first to Sb3+ and then to Sb2S3. The Sb2S3 and SnS2 are then dissolved as chloro-complexes in hydrochloric acid and their presence is confirmed by appropriate tests. The confirmatory test for tin takes advantage of the two oxidation states, +2 and +4, of the metal. Aluminum is added to reduce Sn4+ to Sn2+: 2 Al(s) + 3 Sn4+ (aq) → 2 Al3+ (aq) + 3 Sn2+ (aq)

(6)

The presence of Sn2+ in this solution is detected by adding mercury(II) chloride, HgCl2. This brings about another oxidation-reduction reaction: Sn2+ (aq) + 2 Hg2+ (aq) + 2 Cl− (aq) → Sn4+ (aq) + Hg2Cl2 (s) white

(7)


294

Experiment 37

Qualitative Analysis of Group II Cations

Formation of a white precipitate of insoluble Hg2Cl2 confirms the presence of tin. (The precipitate may be grayish because of the formation of finely divided Hg by an oxidation-reduction reaction similar to Reaction 7.) The Sb3+ ion is difficult to confirm in the presence of Sn4+; the colors of the sulfides of these two ions are similar. To prevent interference by Sn4+, the solution to be tested for Sb3+ is first treated with oxalic acid. This forms a very stable oxalato-complex with Sn4+, Sn(C2O4)32−. Treatment with H2S then gives a bright-orange precipitate of Sb2S3 if antimony is present: 2 Sb3+ (aq) + 3 H2S (aq) → Sb2S3(s) + 6 H+ (aq) orange

(8)

As pointed out earlier, CuS and Bi2S3 are insoluble in NaOH solutions, and they do not dissolve in hydrochloric acid. However, these two sulfides can be brought into solution by treatment with the oxidizing acid, HNO3. The reactions that occur are of the oxidation-reduction type. The NO3− ion is reduced, usually to NO2; S2− ions are oxidized to elementary sulfur and the cation, Cu2+ or Bi3+, is brought into solution. The equations are: CuS(s) + 4 H+ (aq) + 2 NO3− (aq) → Cu2+ (aq) + S(s) + 2 NO2(g) + 2 H2O

(9)

Bi2S3(s) + 12 H+ (aq) + 6 NO3− (aq) → 2 Bi3+ (aq) + 3 S(s) + 6 NO2(g) + 6 H2O

(10)

The two ions, Cu2+ and Bi3+, are easily separated by the addition of aqueous ammonia. The Cu2+ ion is converted to the deep-blue complex, Cu(NH3)42+: Cu2+ (aq) + 4 NH3 (aq) → Cu(NH3)42+ (aq) deep blue

(11)

The reaction of ammonia with Bi3+ is quite different. The OH− ions produced by the reaction of NH3 with water precipitate Bi3+ as Bi(OH)3. We can consider that the reaction occurs in two steps: 3 NH3 (aq) + 3 H2O → 3 NH4+ (aq) + 3 OH− (aq) ← Bi3+ (aq ) + 3 OH − (aq ) → Bi(OH)3 (s) Bi3+ (aq ) + 3 NH3 (aq ) + 3 H 2O → Bi(OH)3 (s) + 3 NH 4 + (aq ) white

(12)

To confirm the presence of Bi3+, the precipitate of Bi(OH)3 is dissolved by treating with hydrochloric acid: Bi(OH)3(s) + 3 H+ (aq) → Bi3+ (aq) + 3 H2O

(13)

The solution formed is poured into distilled water. If Bi3+ is present, a white precipitate of bismuth oxychloride, BiOCl, will form: Bi3+ (aq) + H2O + Cl− (aq) → BiOCl(s) + 2 H+ (aq) white

(14)

Experimental Procedure Step 1.

Adjustment of pH Prior to Precipitation. Pour a 1-mL sample of the “known” solution for

Group II, containing equal volumes of 0.1 M solutions of the nitrates or chlorides of Sn4+, Sb3+, Cu2+, and Bi3+, into a small test tube. Prepare a boiling-water bath, using a 250-mL beaker about 2/3 full of water. Use another beaker full of water as the storage and rinsing place for your stirring rods. Prepare a pH test paper by making ten spots of methyl violet indicator on a piece of filter paper, about one drop to a spot. Let the paper dry in the air. The Group II known will be very acidic because of the presence of HCl, which is necessary to keep the salts of Bi3+, Sn4+, and Sb3+ in solution. Add 6 M NH3, drop by drop, until the solution, after stirring, produces a violet spot on the pH test paper; test for pH


Experiment 37


295

by dipping your stirring rod in the solution and touching it to the paper. (A precipitate will probably form during this step because of the formation of insoluble salts of the Group II cations.) Then add 1 drop of 6 M HCl for each milliliter of solution. This should bring the pH of the solution to about 0.5. Test the pH, using the test paper. At a pH of 0.5, methyl violet will have a blue-green color. Compare your spot with that made by putting a drop of 0.3 M HCl on the test paper (pH = 0.5). Adjust the pH of your solution as necessary by adding HCl or NH3, until the pH test gives about the right color on the paper. If you have trouble deciding on the color, centrifuging out the precipitate may help. When you have established the proper pH, add 1 mL 1 M thioacetamide to the solution and stir. Step 2.

Precipitation of the Group II Sulfides. Heat the test tube in the boiling water bath for

at least five minutes. C A U T I O N : Small amounts of H2S will be liberated; this gas is toxic, so avoid inhaling it unnecessarily. In the presence of Group II ions, a precipitate will form: typically, its color will be initially light, gradually darkening, and finally becoming black. Continue to heat the tube for at least two minutes after the color has stopped changing. Cool the test tube under the water tap and let it stand for a minute or so. Centrifuge out the precipitate and decant the solution, which will contain any Group III ions if they are present, into a test tube. Test the solution for completeness of precipitation by adding two drops of thioacetamide and letting it stand for a minute. If a precipitate forms, add a few drops of thioacetamide and heat again in the water bath. Combine the two batches of precipitate. Wash the precipitate with 2 mL 1 M NH4Cl solution; stir thoroughly. Centrifuge and discard the wash into a waste beaker. Step 3.

Separation of the Group II Sulfides into Two Subgroups. To the precipitate from

Step 2, add 2 mL 1 M NaOH. Heat in the water bath, with stirring, for two minutes. Any SnS2 or Sb2S3 should dissolve. The residue will typically be dark and may contain CuS and Bi2S3. Centrifuge and decant the yellow liquid into a test tube. Wash the precipitate twice with 2 mL water and a few drops of 1 M NaOH. Stir, centrifuge, and decant, discarding the liquid each time into a waste beaker. To the precipitate add 2 mL 6 M HNO3 and put the test tube aside (Step 8). At this point the tin and antimony are present in the solution as complex ions, and the copper and bismuth are in the sulfide precipitate. Step 4.

Reprecipitation of SnS2 and Sb2S3. To the yellow liquid from Step 3 add 6 M HCl drop

by drop until the mixture is just acidic to litmus. Upon acidification, the tin and antimony will again precipitate as orange sulfides. Add five drops of 1 M thioacetamide and heat in the water bath for two minutes to complete the precipitation. Centrifuge, and decant the liquid, which may be discarded. Step 5.

Dissolving SnS2 and Sb2S3 in Acid Solution. Add 2 mL 6 M HCl to the precipitate from

Step 4, and heat in the water bath for a minute or two to dissolve the precipitate. Transfer the solution to a 30-mL beaker, and boil it, gently, for about a minute, to drive out the H2S. Add 1 mL 6 M HCl and 1 mL water and pour the liquid back into a small test tube. If there is an insoluble residue, centrifuge it out and decant the liquid into a test tube. Step 6.

Confirmation of the Presence of Tin. Pour half of the solution from Step 5 into a test

tube and add 2 mL 6 M HCl and a 1-cm length of 24-gauge aluminum wire. Heat the test tube in the water bath to promote reaction of the Al and production of H2. In this reducing medium, any tin present will be converted to Sn2+ and any antimony to the metal, which will appear as black specks. Heat the tube for two minutes after all of the wire has reacted. Centrifuge out any solid and decant the liquid into a test tube. To the liquid add a drop or two of 0.1 M HgCl2. A white or gray cloudiness, produced as Hg2Cl2 or Hg slowly forms, establishes the presence of tin. Step 7.

Confirmation of the Presence of Antimony. To the other half of the solution from

Step 5, add 1 M NaOH to bring the pH to 0.5. Add 2 mL water and about 0.5 g of oxalic acid, and stir until no more crystals dissolve. Oxalic acid forms a very stable complex


296

Experiment 37


with the Sn4+ ion. Add 1 mL 1 M thioacetamide and put the test tube in the water bath. The formation of a red-orange precipitate of Sb2S3 confirms the presence of antimony. Step 8.

Dissolving the CuS and Bi2S3. Heat the test tube containing the precipitate from Step 3

in the hot water bath. Any sulfides that have not already dissolved should go into solution in a minute or two, possibly leaving some insoluble sulfur residue. Continue heating until no further reaction appears to occur, at least two minutes after the initial changes. Centrifuge and decant the solution, which may contain Cu2+ and Bi3+, into a test tube. Discard the residue. Step 9.

Confirmation of the Presence of Copper. To the solution from Step 8 add 6 M NH3

dropwise until the mixture is just basic to litmus. Add 0.5 mL more. Centrifuge out any white precipitate that forms, and decant the liquid into a test tube. If the liquid is deep blue, the color is due to the Cu(NH3)42+ ion, and copper is present. Step 10.

Confirmation of the Presence of Bismuth. Wash the precipitate from Step 9, which

probably contains bismuth, with 1 mL water and 0.5 mL 6 M NH3. Stir, centrifuge, and discard the wash. To the precipitate add 0.5 mL 6 M HCl and 0.5 mL water. Stir to dissolve any Bi(OH)3 that is present. Add the solution drop by drop, to 400 mL water in a 600-mL beaker. A white cloudiness, caused by slow precipitation of BiOCl, confirms the presence of bismuth. Step 11.

When you have completed the analysis of your known, obtain a Group II unknown and test it for the presence of Sn4+, Sb3+, Cu2+, and Bi3+.

Optional Experiment: Establish the presence of copper ion in a mineral supplement tablet. Estimate the amount of copper present by comparison with a suitable standard. As you complete each part of the experiment, put the waste products in a beaker. At the end of the experiment, pour the contents of the beaker into a waste crock, unless directed otherwise by your instructor.



Name ____________________________________

Section ________________________________

Experiment 37

Observations and Report Sheet: Qualitative Analysis of Group II Cations Flow Diagram for Group II

Observations on known (record on diagram if different from those on prepared diagram). Observations on unknown (record on diagram in colored pencil to distinguish from observations on known). Unknown no. ____________ Ions reported present ____________ ____________ ____________ ____________



Name ____________________________________

Section ________________________________

Experiment 37

Advance Study Assignment: Qualitative Analysis of Group II Cations 1. Prepare a complete flow diagram for the separation and identification of the Group II cations and put it on your report sheet on page 297. 2. Write balanced net ionic equations for the following reactions: a. Precipitation of the tin(IV) sulfide with H2S.

b. The confirmatory test for antimony.

c. The confirmatory test for bismuth.

d. The dissolving of Bi2S3 in hot nitric acid.

3. A solution that may contain Cu2+, Bi3+, Sn4+, or Sb3+ ions is treated with thioacetamide in an acid medium. The black precipitate that forms is partly soluble in strongly alkaline solution. The precipitate that remains is soluble in 6 M HNO3 and gives only a blue solution on treatment with excess NH3. The alkaline solution, when acidified, produces an orange precipitate. On the basis of this information, which ions are present, which are absent, and which are still in doubt? (As with Group I, evidence other than confirmatory tests may show the presence or absence of an ion.) Present ____________ Absent ____________ In doubt ____________



Experiment 38 Qualitative Analysis of Group III Cations

Precipitation and Separation of the Group III Ions Four ions in Group III are Cr3+, Al3+, Fe3+, and Ni2+. The first step in the analysis involves treating the solution with sodium hydroxide, NaOH, and sodium hypochlorite, NaOCl. The OCl− ion oxidizes Cr3+ to CrO42−: 2 Cr3+ (aq) + 3 OCl− (aq) + 10 OH− (aq) → 2 CrO42− (aq) + 3 Cl− (aq) + 5 H2O yellow

(1)

The chromate ion, CrO42−, stays in solution. The same is true of the hydroxo-complex ion Al(OH)4−, formed by the reaction of Al3+ with excess OH−: Al3+ (aq) + 4 OH− (aq) → Al(OH)4− (aq)

(2)

In contrast, the other two ions in the group form insoluble hydroxides under these conditions: Ni2+ (aq) + 2 OH− (aq) → Ni(OH)2(s) green

(3)

Fe3+ (aq) + 3 OH− (aq) → Fe(OH)3(s) red

(4)

The Ni2+ and Fe3+ ions, unlike Al3+, do not readily form hydroxo-complexes. Unlike Cr3+, they do not have a stable higher oxidation state and so are not oxidized by OCl−. To separate aluminum from chromium, the solution containing CrO42− and Al(OH)4− is first acidified. This destroys the hydroxo-complex of aluminum: Al(OH)4− (aq) + 4 H+ (aq) → Al3+ (aq) + 4 H2O

(5)

Treatment with aqueous ammonia then gives a white gelatinous precipitate of aluminum hydroxide. We can think of this reaction as occurring in two steps: 3 NH3 (aq) + 3 H2O

NH4+ (aq) + 3 OH− (aq)

Al3+ (aq ) + 3 OH − (aq ) → Al(OH)3 (s) Al (aq ) + 3 NH3 (aq ) + 3 H 2O → Al(OH)3 (s) + 3 NH 4 + (aq ) white 3+

(6)

The concentration of OH− in dilute NH3 is too low to form the Al(OH)4− complex ion by Reaction 2. The CrO42− ion remains in solution after Al3+ has been precipitated. It can be tested for by precipitation as yellow BaCrO4 by the addition of BaCl2 solution: Ba2+ (aq) + CrO42− (aq) → BaCrO4(s) light yellow

(7)

The precipitate of BaCrO4 is dissolved in acid; the solution formed is then treated with hydrogen peroxide, H2O2. A deep-blue color is produced, because of the presence of a peroxo-compound, probably CrO5. The reaction may be represented by the overall equation: 2 BaCrO4(s) + 4 H+ (aq) + 4 H2O2 (aq) → 2 Ba2+ (aq) + 2 CrO5 (aq) + 6 H2O blue

(8)


302

Experiment 38

Qualitative Analysis of Group III Cations

The mixed precipitate of Ni(OH)2 and Fe(OH)3 formed by Reactions 3 and 4 is dissolved by adding a strong acid, HNO3. An acid-base reaction occurs: Ni(OH)2(s) + 2 H+ (aq) → Ni2+ (aq) + 2 H2O

(9)

Fe(OH)3(s) + 3 H+ (aq) → Fe3+ (aq) + 3 H2O

(10)

At this point, the Ni2+ and Fe3+ ions are separated by adding ammonia. The Ni2+ ion is converted to the deepblue complex Ni(NH3)62+, which stays in solution: Ni2+ (aq) + 6 NH3 (aq) → Ni(NH3)62+ (aq) blue

(11)

while the Fe3+ ion, which does not readily form a complex with NH3, is reprecipitated as Fe(OH)3: 3 NH3 (aq) + 3 H2O 3 NH4+ (aq) + 3 OH− (aq) 3+ − Fe (aq ) + 3 OH (aq ) → Fe(OH)3 (s) 3+ Fe (aq ) + 3 NH3 (aq ) + 3 H 2O → Fe(OH)3 (s) + 3 NH 4 + (aq )

(12)

red The confirmatory test for nickel in the solution is made by adding an organic reagent, dimethylglyoxime, C4H8N2O2. This gives a deep rose-colored precipitate with nickel: Ni2+ (aq) + 2 C4H8N2O2 (aq) → Ni(C4H7N2O2)2(s) + 2 H+ (aq) rose

(13)

We can confirm Fe3+ by dissolving the precipitate of Fe(OH)3 in HCl (Reaction 10) and adding KSCN solution. If iron(III) is present, the blood-red FeSCN2+ complex ion will form: Fe3+ (aq) + SCN− (aq) → FeSCN2+ (aq) red

(14)

Experimental Procedure Step 1. If you are testing a solution from which the Group II ions have been precipitated

(Experiment 37), remove the excess H2S and excess acid by boiling the solution until the volume is reduced to about 1 mL. Remove any sulfur residue by centrifuging the solution. If you are working on the analysis of Group III cations only, prepare a known solution containing Fe3+, Al3+, Cr3+, and Ni2+ by mixing together 0.5-mL portions of each of the appropriate 0.1 M solutions containing those cations. Step 2. Oxidation of Cr(III) to Cr(VI) and Separation of Insoluble Hydroxides. Add 1 mL 6 M

NaOH to 1 mL of the known solution in a 30-mL beaker. Boil very gently for a minute, stirring to minimize bumping. Remove heat, and slowly add 1 mL 1 M NaOCl, sodium hypochlorite. Swirl the beaker for 30 seconds, using your tongs if necessary. Then boil the mixture gently for a minute. Add 0.5 mL 6 M NH3 and let stand for 30 seconds. Then boil for another minute. Transfer the mixture to a test tube and centrifuge out the solid, which contains iron and nickel hydroxides. Decant the solution, which contains chromium and aluminum (CrO42− and Al(OH)4−) ions, into a test tube (Step 3). Wash the solid twice with 2 mL water and 0.5 mL 6 M NaOH; after mixing, centrifuge each time, discarding the wash. Add 1 mL water and 1 mL 6 M H2SO4 to the solid and put the test tube aside (Step 6). Step 3. Separation of Al from Cr. Acidify the solution from Step 2 by adding 6 M acetic acid

slowly until, after stirring, the mixture is definitely acidic to litmus. If necessary, transfer the solution to a 50-mL beaker and boil it to reduce its volume to about 3 mL. Pour the solution into a test tube. Add 6 M NH3, drop by drop, until the solution is basic to litmus, and then


Experiment 38

Qualitative Analysis of Group III Cations

303

add 0.5 mL in excess. Sir the mixture for a minute or so to bring the system to equilibrium. If aluminum is present, a light, translucent, gelatinous white precipitate of Al(OH)3 should be floating in the clear (possibly yellow) solution. Centrifuge out the solid, and transfer the liquid, which may contain CrO42−, into a test tube (Step 5). Step 4. Confirmation of the Presence of Aluminum. Wash the precipitate from Step 3 with 3 mL

water once or twice, while warming the test tube in the water bath and stirring well. Centrifuge and discard the wash. Dissolve the precipitate in 2 drops 6 M HC2H3O2, acetic acid, no more, no less. Add 3 mL water and 2 drops catechol violet reagent and stir. If Al3+ is present, the solution will turn blue. Step 5. Confirmation of the Presence of Chromium. If the solution from Step 3 is yellow,

chromium is probably present; if it is colorless, chromium is absent. To the solution add 0.5 mL 1 M BaCl2. In the presence of chromium you obtain a finely divided yellow precipitate of BaCrO4, which may be mixed with a white precipitate of BaSO4. Put the test tube in the boiling water bath for a few minutes; then centrifuge out the solid and discard the liquid. Wash the solid with 2 mL water; centrifuge and discard the wash. To the solid add 0.5 mL 6 M HNO3, and stir to dissolve the BaCrO4. Add 1 mL water, stir the orange solution, and add two drops of 3% H2O2, hydrogen peroxide. A deep blue solution, which may fade quite rapidly, is confirmatory evidence for the presence of chromium. Step 6. Separation of Iron and Nickel. Returning to the precipitate from Step 2, stir to dissolve

the solid in the H2SO4. If necessary, warm the test tube in the water bath to complete the solution process. Then add 6 M NH3 until the solution is basic to litmus. At that point iron will precipitate as brown Fe(OH)3. Add 1 mL more of the NH3, and stir to bring the nickel into solution as Ni(NH3)62+ ion. Centrifuge and decant the liquid into a test tube. Save the precipitate (Step 8). Step 7. Confirmation of the Presence of Nickel. If the solution from Step 6 is blue, nickel is

probably present. To that solution add 0.5 mL dimethylglyoxime reagent. Formation of a rose-red precipitate proves the presence of nickel. Step 8. Confirmation of the Presence of Iron. Dissolve the precipitate from Step 6 in 0.5 mL 6 M

HCl. Add 2 mL water and stir. Then add 2 drops of 0.5 M KSCN. Formation of a deep red solution of FeSCN2+ is a definitive test for iron. Step 9. When you have completed your analysis of the known solution, obtain a Group III unknown

and test it for the possible presence of Fe3+, Al3+, Cr3+, and Ni2+. DISPOSAL OF REACTION PRODUCTS. As you complete each step in the procedure, put the waste products into a beaker. When you are finished with the experiment, pour the contents of the beaker into a waste crock, unless otherwise directed by your instructor.



Name ____________________________________

Section ________________________________

Experiment 38

Observations and Report Sheet: Qualitative Analysis of Group III Cations Flow Diagram for Group III

Observations on known (record on diagram if different from those on prepared diagram). Observations on unknown (record on diagram in colored pencil to distinguish from observations on known). Unknown no. ____________ Ions reported present ____________ ____________ ____________ ____________



Name ____________________________________

Section ________________________________

Experiment 38

Advance Study Assignment: Qualitative Analysis of Group III Cations 1. Prepare a flow diagram for the separation and identification of the ions in Group III and put it on your report sheet. 2. Write balanced net ionic equations for the following reactions: a. Dissolving of Fe(OH)3 in nitric acid.

b. Oxidation of Cr3+ to CrO42− by ClO− in alkaline solution (ClO− is converted to Cl−).

c. The confirmatory test for Ni2+.

d. The confirmatory test for Fe3+.

3. A solution may contain any of the Group III cations. Treatment of the solution with ClO− in alkaline medium yields a yellow solution and a colored precipitate. The acidified solution is unaffected by treatment with NH3. The colored precipitate dissolves in nitric acid; addition of excess NH3 to this acid solution produces only a blue solution. On the basis of this information, which Group III cations are present, absent, or still in doubt? Present ____________ Absent ____________ In doubt ____________



Experiment 39 Identification of a Pure Ionic Solid

n the previous set of experiments we have seen how one can identify which anions, or cations, are actually present in an unknown solution. In this experiment you will be asked to determine the cation and anion that are present in a solid sample of a pure ionic salt. The possible cations will include those we studied in Experiments 36, 37, and 38:

I

Ag+ Cu2+

Pb2+ Bi3+

Sn2+

Hg22+ Cr3+

Al3+

Sb3+

Ni2+

Fe3+

The anions we will consider are those in Experiment 35: CO32−

PO43−

Cl−

SCN−

SO42−

SO32−

C2H3O2−

NO3−

There are 88 possible compounds that you might be given to identify. There are, of course, many ionic compounds that are not included, but the set is reasonably representative. At first sight it might seem that the best approach would be to carry out the procedures for analysis of Groups I, II, and III, in succession, and stop when you get to the cation in your unknown. That attack should work, but it would take longer than necessary, and would require that you have a solution of the compound to work with. Given a knowledge of the properties of the possible compounds, an experienced chemist would do some preliminary tests to find out the solubility properties of the sample and try to use those to narrow down very significantly the number of compounds she needed to consider. The solubility in various solvents, plus color, and odor if any, should afford some very useful information. Most of the possible compounds are not soluble in water, but may dissolve in strong acids, like 6 M HCl, or strong bases, like 6 M NaOH, or solutions containing ligands that form stable complex ions with the cation in the compound. Some important complexing ligands that may dissolve otherwise insoluble solids containing the cations we are considering include: NH3

OH−

Cl−

C2H3O2−

Nearly every compound in our set of 88 will dissolve in, or react with, at least one of the above species in solution. Before beginning the analysis, it will be helpful to note some of the properties of the cations that distinguish them from one another. We have given those properties in Appendix IIA. You should read that Appendix and complete the Advance Study Assignment before coming to lab. You will need the information in the Appendix in your analysis.

Experimental Procedure In this experiment you will be given two unknowns to identify. The cation in your first sample will be colored, or will produce colored solutions. The second unknown will be colorless and somewhat more difficult to identify. In both cases the analysis will involve determining the solubility of the solid in water and some acidic or basic solutions. You should have about a half-gram of finely divided solid available. Read the section on qualitative analysis at the end of Appendix IV if you haven’t already. Prepare a boiling-water bath and a bath for rinsing your stirring rods.


310

Experiment 39

Identification of a Pure Ionic Solid

Identifying the Cation in a Colored Ionic Solid The solvents we will consider are the following: 1. Water

2. 6 M HNO3

3. 6 M HCl

4. 6 M NaOH

5. 6 M NH3

6. 6 M H2SO4

Determine the solubility behavior of the solid in as many of the solvents as you need to find one in which the solid will dissolve. Use a small sample, about 50 mg, enough to cover a 2-mm circle on the end of a small spatula. Put the solid in a small test tube and add about 1 mL of water. Stir the solid with a stirring rod for a minute or so, noting any color changes of the solid or the solution. If the solid does not dissolve, put the tube in your boiling-water bath for a few minutes. If at this point you have a solution, add 5 mL more water and stir. From the color of the solution you should be able to make a good guess as to which cation is in your unknown. (The information in Appendix II or IIA should be helpful.) Use 1 mL of your solution to carry out the confirmatory test for that cation, as directed in Experiment 37 or 38. Make sure that you set the pH and any other conditions for the test properly before making a decision. If your guess is confirmed, you may proceed to analysis for the anion. If you were wrong, perform the confirmatory test for another cation that seems likely. It should not be difficult to identify the cation that is present. If your sample does not dissolve in water, try 6 M HNO3 as a solvent, using the same procedure as with water (about 1 mL of reagent and 50 mg of solid). If the sample goes into solution, in the cold, or in the water bath, add 5 mL of water, stir, and proceed to identify the cation present. With HNO3 you may get some effervescence. This indicates that the anion is either carbonate or sulfite; with the former, odorless CO2, is given off, while with the latter, SO2, with its characteristic sharp odor, is evolved. Nearly all the possible compounds in our set that are colored are also soluble in either hot water or hot HNO3. If you by chance have one that will not dissolve in those reagents, make the solubility test using 6 M HCl as the solvent. That should work if the first two solvents have not. In this part of the experiment you should not need to use the last three solvents on the list.

Anion Analysis Before beginning analysis for the anion in your unknown, you should find it helpful to know that some anions are much more common than others in the chemicals that are ordinarily available in the stockroom. It is easiest to work with chemicals that are soluble in water. Nitrates, chlorides, sulfates, and acetates are often soluble, and are the most common salts we find on the shelves. Thiocyanates and sulfites are sometimes soluble, but are usually found as sodium or potassium salts, and so are not likely in the compounds in the set we are using. Carbonates and phosphates are insoluble in water, and occasionally are available. Hydroxides and oxides are commonly found in stockrooms, since they are easy to synthesize. They are insoluble, and will not be in our unknowns unless your instructor tells you otherwise. Given the above information and the spot tests in Experiment 35, determine which anion is in your solid. For each test, use about 1 or 2 mL of the solution you prepared. You don’t need to do all the spot tests, only those that seem reasonable in light of the previous discussion. If you used water as your solvent, any positive test for an anion should be conclusive. If you used HNO3, you can’t test for nitrate ion in that solution. Similarly you can’t test for chloride ion in an HCl solution. Some anions may be difficult to determine by the usual spot test, since the cation present may interfere. That is the case with acetate and nitrate, where the test solution is basic and will cause the cation to precipitate. With acetates, a simple test is to add 1 mL 6 M H2SO4 to 50 mg of the solid, as if you were testing for solubility. Heat the solution, or slurry, in the hot-water bath. If the sample contains acetate, acetic acid will be volatilized and can be detected by its characteristic odor. Put your stirring rod in the warm liquid, and, cautiously, sniff the rod. If there are no interfering anions, the odor of vinegar is an excellent test for acetate. With nitrates, begin by adding 1 mL 6 M NaOH to a small sample of solid unknown. If you get a hydroxide precipitate, you can try simply proceeding with the spot test. If the cation forms a hydroxide complex, you can try the spot test, and it may work. In many cases, the aluminum metal will reduce the cation to the metal, and you will get a black solid. With excess Al, you still may be able to detect evolved NH3 and establish the


Experiment 39


311

presence of nitrate. The test for nitrate is perhaps the most difficult, and if worst comes to worst, you can treat your acidic solution of unknown with excess 1 M Na2CO3 until effervescence ceases and the solution is definitely basic; add 0.5 mL more. Centrifuge out the solid cation-containing carbonate, which can be discarded. Acidify the decanted liquid with 6 M H2SO4 drop by drop until CO2 evolution ceases, and the solution is acidic. Carefully boil this solution down to about 2 mL and use it for the nitrate test and any other anion tests that seem indicated.

Analysis of a Colorless Ionic Solid As your second unknown to analyze you will be given a white solid, which may have very limited solubility in water. It may also be insoluble in some of the acids and bases on our original list. The procedure initially at least is the same as you used with the colored sample. Check the solubility of the solid in water, both at room temperature and in the water bath. Some of the solids do dissolve. With these, and the rest of the unknowns, the solubility in the other solvents is likely to furnish clues about the cation that is present. So, carry out solubility tests with all six of the solvents we listed (again, about 1 mL of reagent added to 50 mg of solid). Before deciding on the solubility in a given solvent, make sure you give the solid time to dissolve if it is going to, with stirring, particularly in the hot-water bath. Some solids have been in storage for years and are slow to get their act together. If the solid does not dissolve, add 1 mL water and stir; that may help. Solids containing cations that form complexes with Cl− or OH− are likely to dissolve in 6 M HCl or 6 NaOH, respectively. Those with cations that form ammonia complexes will tend to dissolve in 6 M NH3. The solubility depends on the Ksp of the solid and the stability of the complex, so a complexing ligand may dissolve one solid and not another. Nitrate ion and sulfate ion are not very good complexing ions, so 6 M HNO3 and 6 M H2SO4 will not dissolve solids by complex ion formation. If they dissolve a solid, it is by reacting with an anion in the solid to form a weak acid or by reacting with a basic salt like BiOCl. There are a few compounds in our possible set of unknowns that will not dissolve in any of the six reagents, or may dissolve in only one reagent. Those compounds are all discussed in Appendix IIA. When you have completed the solubility tests on your solid, compare your results with those you obtained in your Advance Study Assignment from information in Appendix IIA. You should be able to narrow down the list of possible cations to only a few, and in some cases you will be able to decide not only which cation is present but also which anion. If you can dissolve the solid in water or in acid, you can proceed to carry out a confirmatory test for any likely cations, as described in Experiments 36, 37, and 38. With the colorless unknowns the number of likely anions is even less than with the first unknown, with nitrate and chloride being most common. If the sample will dissolve in water, or in HNO3 or HCl, you can carry out spot tests for any anions that are not present in the solvent. For some solids, you will need to identify the anion on the basis of the solubility behavior of the salt. On the report page note all your observations and the name and formula of the compound you believe is in your unknown. When you are finished with the experiment, discard all the reaction products in your 250-mL beaker in the waste crock.




Name ____________________________________

Section ________________________________

Experiment 39

Observations and Report Sheet: Identification of a Pure Ionic Solid Color of first unknown ____________ Solubility properties of unknown _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ Cations that are likely components of unknown ____________ ____________ Confirmatory tests performed to identify cation Procedures used _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ Results observed _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ ____________ is present Spot tests performed to identify anion Procedures used _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ Results observed _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ ____________ is present Formula of colored compound

____________


314

Experiment 39


Solubility behavior of white unknown Water

_____________________________

6 M HNO3 _____________________________

6 M HCl _____________________________

6 M NaOH _____________________________

6 M NH3 _____________________________

6 M H2SO4 _____________________________

Reasoning and results of confirmatory tests _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ Formula of white compound ____________


Name ____________________________________

Section ________________________________

Experiment 39

Advance Study Assignment: Identification of a Pure Ionic Solid Use the information in Appendix II and Appendix IIA to find answers to the questions. 1. What are the formulas and colors of the colored cations used in this experiment? _________________________________________________________________________ 2. Which of the colorless cations A. will form complex ions with NH3?

____________

B. will form complex ions with OH−?

____________

C. will form complex ions with Cl−?

____________

D. have colored, rather than black, sulfides?

____________

E. have one or more water-soluble salts?

____________

3. State the formula of a colorless compound that A. dissolves in 6 M NH3.

____________

B. turns black when NH3 is added.

____________

C. is much more soluble in hot water than in cold.

____________

D. is soluble in 6 M NaOH but not in 6 M H2SO4 .

____________

E. is not soluble in any of the solvents we use.

____________

4. A colorless ionic solid is insoluble in water, and in 6 M NH3, but will dissolve in 6 M HCl and 6 M NaOH. Name two compounds that would have these properties. ____________ and ____________



Experiment 40 The Ten Test Tube Mystery

aving carried out at least some of the previous experiments on qualitative analysis, you should have acquired some familiarity with the properties of the cations and anions you studied. Along with this you should at this point have had some experience with interpretation of the data in Appendix II on the solubilities of ionic substances in different media. In this experiment we are going to ask you to apply what you have learned to a related but somewhat different kind of problem. You will be furnished ten numbered test tubes in each of which there will be a solution of a single substance. You will be provided, one week in advance, with a list giving the formula and molarity of each of the ten solutes that will be used. Your problem in the laboratory will be to find out which solution is in which test tube, that is, to assign a test tube number to each of the solution compositions. You are to do this by intermixing small volumes of the solutions in the test tubes. No external reagents or acid-base indicators such as litmus are allowed. You are permitted, however, to use the odor and color of the different solute species and to make use of heat effects in reactions in your system for identification. Each student will have a different set of test tube number-solution composition correlations, and there will be several different sets of compositions as well. Of the ten solutions, four are common laboratory reagents. They are 6 M HCl, 3 M H2SO4, 6 M NH3, and 6 M NaOH. The other solutions will usually be 0.1 M nitrate, chloride, or sulfate solutions of the cations, which have been studied in previous experiments in this manual (Experiments 12, 36, 37, and 38). To determine which solution is in each test tube, you will need to know what happens when the various solutions are mixed, one with another. In some cases, nothing happens that you can observe. This will often be the case when a solution containing one of the cations is mixed with a solution of another. When one of the reagents is mixed with a cation solution you may get a precipitate, white or colored, and that precipitate may dissolve in excess reagent by complex ion formation. In a few cases a gas may be evolved. When one laboratory reagent is mixed with another, you may find that the resulting solution gets very hot and/or that a visible vapor is produced. There is no way that you will be able to solve your particular test tube mystery without doing some preliminary work. You will need to know what to expect when any two of your ten solutions are mixed. You can find this out, for any pair, by scrutiny of Appendix II, by consulting your chemistry text, and by referring to various reference works on qualitative analysis.* A convenient way to tabulate the information you obtain is to set up a matrix with ten columns and ten rows, one for each solution. The key information about a mixture of two solutions is put in the space where the row for one solution and the column for the other intersect (as is done in Appendix II for various cations and anions). In that space you might put an NR, to indicate no apparent reaction on mixing of the two solutions. If a precipitate forms, put a P, followed by a D if the precipitate dissolves in excess reagent. If the precipitate or final solution is colored, state the color. If heat is evolved, write an H; if a gas or smoke is formed, put a G or S. Since mixing solution A with B is the same as mixing B with A, not all 100 spaces in the 10-by-10 matrix need to be filled. Actually, there are only 45 possible different pairs, since A with A is not very informative either. Because you are allowed to use the odor or color of a solution to identify it, the problem is somewhat simpler than it might first appear. In each set of ten solutions you will probably be able to identify at least two solutions by odor and color tests. Knowing those solutions, you can make mixtures with the other solutions in which one of the components is known. From the results obtained with those mixtures, and the information in

H

*For example, the following references may be useful: 1 Qualitative Analysis and the Properties of Ions in Aqueous Solutions. 2nd edition by E. J. Slowinski and W. L. Masterton, Saunders College Publishing, 1990. 2

Handbook of Chemistry and Physics. Chemical Rubber Publishing, 2010.


318

Experiment 40

The Ten Test Tube Mystery

the matrix, you can identify other solutions. These can be used to identify still others, until finally the entire set of ten is identified unequivocally. Let us now go through the various steps that would be involved in the solution to a somewhat simpler problem than the one you will solve. There are several steps, including constructing the reaction matrix, identifying solutions by simple observations, and a rationale for efficient mixing tests. Let us assume we have to identify the following six solutions in numbered test tubes: 0.1 M NiSO4

0.1 M BiCl3(in 3 M HCl)

0.1 M BaCl2

6 M NaOH

0.1 M Al(NO3)3

3 M H2SO4

Construction of Reaction Matrix In the matrix there will be six rows and six columns, one for each of the solutions. The matrix would be set up as in Table 40.1. Each solution contains a cation and an anion, both of which need to be considered. We need to include every possible pair of solutions: for six solutions there are 15 possible pairs. So we need 15 pieces of information as to what happens when a pair of solutions is mixed. The first pair we might consider is Al(NO3)3 and BaCl2, containing Al3+, NO3−, Ba2+, and Cl− ions. Consulting Appendix II, we see that in the Al3+, Cl− space there is an S, meaning that AlCl3 is soluble and will not precipitate when aluminum and chloride ions are mixed. Similarly, Ba(NO3)3 is soluble because all nitrates are soluble. This means that there will be no reaction when the solutions of Al(NO3)3 and BaCl2 are mixed. So, in the space for Al(NO3)3-BaCl2, we have written NR. The same is true if we mix solutions of Al(NO3)3 and NiSO4, so we insert NR in that space. Since BiCl3 won’t react with Al(NO3)3 solutions, there is an NR in that space too. However, when Al(NO3)3 is mixed with NaOH, one of the possible products is Al(OH)3, which we see in Appendix II is insoluble in water, but dissolves in acid and excess OH− ion (A, B). This means that on adding NaOH to Al(NO3)3, we would initially get a precipitate of Al(OH)3, P, but that it would dissolve in excess NaOH, D. The precipitate is white and the solution is colorless, so we have just a P and a D in the space for that mixture. Proceeding now to the BaCl2 row, we don’t need to consider the first two columns, because they would give no new information. If BaCl2 and NiSO4 were mixed, BaSO4 is a possible product. In Appendix II we see that BaSO4 is insoluble in all common solvents (I). Hence, on mixing those solutions, we would get a precipitate of BaSO4, P. Using the Handbook of Chemistry and Physics, or some other source, we find that it is white, so a P goes into that space. With BiCl3 there would be no reaction; with NaOH we might get a slight precipitate (S− for Ba(OH)2 in Appendix II), so we put a P in the appropriate space. With H2SO4, we would again expect to get a white precipitate of BaSO4, P. In the row for NiSO4 we note that the color of the solution is green, since in Appendix II we see that Ni2+(aq) ion is green. The first column is that for BiCl3; no precipitate forms on mixing solutions of NiSO4 and BiCl3, so NR is in that space. With NaOH, Ni(OH)2 would precipitate, since the entry in Appendix II for Ni2+, OH−, is not S. From the Handbook or other sources we find that the precipitate is green, so in the space we put a P (green). In Appendix II the Ni(OH)2 entry is A, N, which tells us that the precipitate will dissolve Table 40.1 Reaction Matrix for Six Solutions Al(NO3)3

Al(NO3)3 BaCl2 NiSO4(green) BiCl3 NaOH

BaCl2

NiSO4

BiCl3

NaOH

H2SO4

NR

NR P

NR NR NR

P, D sl P P (green) P, H

NR P NR NR H

H2SO4


Experiment 40

The Ten Test Tube Mystery

319

in 6 M NH3 (but not in 6 M NaOH since there is no B in the space). There would be no reaction between solutions of NiSO4 and H2SO4(NR). With BiCl3 in 3 M HCl we have both a salt and an acid in the solution. Addition of 6 M NaOH should produce a precipitate (white) of Bi(OH)3, P. There would also be a substantial exothermic acid-base reaction between H+ ions in the HCl and OH− ions in the NaOH, so there would be a very noticeable rise in temperature on mixing, as denoted by H in the space. There would be no reaction with H2SO4. When 6 M NaOH is mixed with 3 M H2SO4 there should be a large heat effect because of the acid-base reaction that occurs, so in the space for NaOH-H2SO4 we have an H. The matrix in Table 40.1 summarizes the reaction information that will be needed for identification of the solutions in the test tubes. Your matrix can be constructed by the reasoning used in making this one.

Identifying Solutions by Simple Observations When we made the matrix for the six test tube system, we noted that the NiSO4 solution would be green, since Ni2+ ion is green. So, as soon as we see the six text tubes, we can pick out the 0.1 M NiSO4. Let us assume that it is in Test Tube 4. We have now identified one solution out of six.

Selection of Efficient Mixing Tests Knowing that Test Tube 4 contains NiSO4, we mix the solution in Test Tube 4 with all of the other solutions. For each mixture, we record what we observe. The results might be as follows: 4+1 4+2 4+3 4+5 4+6

no obvious reaction no obvious reaction green precipitate forms no obvious reaction white precipitate

Referring to the matrix, we would expect precipitates for mixtures of NiSO4 and NaOH, and NiSO4 and BaCl2. The former precipitate would be green, the latter white. Clearly the solution in Test Tube 3 is 6 M NaOH. The solution in Test Tube 6 must be 0.1 M BaCl2. At this point three solutions have been identified. Because 6 M NaOH reacts with several of the solutions, and because we know now that it is in Test Tube 3, we mix the solution in Test Tube 3 with those that remain unidentified, with the following results: 3+1 3+2 3+5

solution becomes hot white precipitate, solution gets hot apparently no reaction

Again, referring to the matrix we would expect heat evolution with mixtures of NaOH and H2SO4, and NaOH and the HCl in the BiCl3. In addition, we would get a white precipitate with BiCl3. It is obvious that Solution 1 is 3 M H2SO4. Solution 2 must be 0.1 M BiCl3 in 3 M HCl. By elimination Solution 5 must be Al(NO3)3. We repeat the 3 + 5 mixture, since there should have been an initial precipitate that dissolved in excess NaOH. This time, using one drop of Solution 3 added to 1 mL of Solution 5, we get a precipitate. In excess of Solution 3 the precipitate dissolves as the aluminum hydroxide complex ion is formed.

Experimental Procedure Using the approach we have outlined in our example, carry out tests as necessary for the identification of the solutions in your set. After you have tentatively assigned each solution to its test tube, carry out some confirmatory mixing reactions, so that you have at least two mixtures that give results that support your conclusions. DISPOSAL OF REACTION PRODUCTS.

Dispose of all reaction products in the waste crock, unless

directed otherwise by your instructor.



Name ____________________________________

Section ________________________________

Experiment 40

Observations and Report Sheet: The Ten Test Tube Mystery Solutions That Were Identified Directly Solution

Observations

Conclusions

Test tube no.

Mixing Tests: First Series Test tube nos.

Observations

Conclusions

Observations

Conclusions

Observations

Conclusions

Mixing Tests: Second Series Test tube nos.

Confirming Tests (total of two per solution) Test tube nos.

Final identifications:

No. 1 ____________

No. 2 ____________

No. 3 ____________

No. 4 ____________

No. 5 ____________

No. 6 ____________

No. 7 ____________

No. 8 ____________

No. 9 ____________

No. 10 ____________

Unknown set no. ____________ Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.


Name ____________________________________

Section ________________________________

Experiment 40

Advance Study Assignment: The Ten Test Tube Mystery 1. Construct the reaction matrix for your set of ten solutions. This should go on the reverse side of this page. This matrix can be made using information in your chemistry text or in Appendix II or, if your instructor allows it, you can go into the lab prior to the scheduled session and work with known solutions to obtain the needed information. 2. Which solutions should you be able to identify by simple observations?

3. Outline the procedure you will follow in identifying the solutions that will require mixing tests. Be as specific as you can about what you will look for and what conclusions you will be able to draw from your observations.



Experiment 41 Preparation of Aspirin

O

ne of the simpler organic reactions that one can carry out is the formation of an ester from an acid and an alcohol: O R

O OH + HO

C an acid

R′ → R

an alcohol

C

R′ + H2O

O

(1)

an ester

In the equation, R and R′ are H atoms or organic fragments like CH3, C2H5, or more complex aromatic groups. There are many esters, because there are many organic acids and alcohols, but they all can be formed, in principle at least, by Reaction 1. The driving force for the reaction is, in general, not very great, so that one ends up with an equilibrium mixture of ester, water, and acid, and alcohol. There are some esters that are solids because of their high molecular weight or other properties. Most of these esters are not soluble in water, so they can be separated from the mixture by crystallization. This experiment deals with an ester of this sort, the substance commonly called aspirin. Aspirin has been used since the beginning of the 20th century to treat headaches and fever. It is the most used medicinal, with a daily production of about 50 tons, about one pill for one out of every two people in this country. Its effectiveness is actually two-fold: as an inflammatory, which explains its painkilling properties; and as an anticoagulant, which is important in its use in prevention of heart attacks and strokes. It works by inhibiting an enzyme that catalyzes formation of precursors of prostaglandins, rather large organic molecules that regulate inflammation of tissues, and of thromboxanes, which stimulate platelet aggregation, the first step in blood clotting. Researchers on prostaglandins have received two Nobel Prizes, the first in 1970, to their discoverer, and the second in 1982, for work relating to the effects of aspirin on prostaglandin behavior. Aspirin can be made by the reaction of the OH group in the salicylic acid molecule with the carboxyl ( COOH) group in acetic acid: OH O

C

O CH3

C

C C

acetic acid

C

O

C

C H

O

H C

OH + HO

OH

→ CH 3 H←

C

H C

O

C

C

C

C H

salicylic acid

C

H

H + H 2O

C H

aspirin (salicylic acid acetate)

A better preparative method, which we will use in this experiment, employs acetic anhydride in the reaction instead of acetic acid. The anhydride can be considered to be the product of a reaction in which two acetic acid molecules combine, with the elimination of a molecule of water. The anhydride will react with the water produced in the esterification reaction and will tend to drive the reaction to the right. A catalyst, normally sulfuric or phosphoric acid, is also used to speed up the reaction.


326

Experiment 41

Preparation of Aspirin

OH O CH3

O

CH3

C

C

O acetic anhydride

H

→ CH3 H←

C

H C

H3 PO4

C C

C

O

C

C

C

O

H C

O + HO

OH

O

C

C

C C

H salicylic acid

O

C

H aspirin

H + CH3

C

OH

C H acetic acid

The aspirin you will prepare in this experiment is relatively impure and should certainly not be taken internally, even if the experiment gives you a bad headache. There are several ways by which the purity of your aspirin can be estimated. Probably the simplest way is to measure its melting point. If the aspirin is pure, it will melt sharply at the literature value of the melting point. If it is impure, the melting point will be lower than the literature value by an amount that is roughly proportional to the amount of impurity present. A more quantitative measure of the purity of your aspirin sample can be obtained by determining the percentage of salicylic acid it contains. Salicylic acid is the most likely impurity in the sample because, unlike acetic acid, it is not very soluble in water. Salicylic acid forms a highly colored magenta complex with Fe(III). By measuring the absorption of light by a solution containing a known amount of aspirin in excess Fe3+ ion, one can easily determine the percentage of salicylic acid present in the aspirin.

Experimental Procedure Weigh a 50-mL Erlenmeyer flask on a triple-beam or top-loading balance and add 2.0 g of salicylic acid. Measure out 5.0 mL of acetic anhydride in your graduated cylinder, and pour it into the flask in such a way as to wash any crystals of salicylic acid on the walls down to the bottom. Add five drops of 85% phosphoric acid to serve as a catalyst. C A U T I O N : Both acetic anhydride and phosphoric acid are reactive chemicals that can give you a bad chemical burn, so use due caution in handling them. If you get any of either on your hands or clothes, wash thoroughly with soap and water. Clamp the flask in place in a beaker of water supported on a wire gauze on a ring stand. Heat the water with a Bunsen burner to about 75°C, stirring the liquid in the flask occasionally with a stirring rod. Maintain this temperature for about 15 minutes, by which time the reaction should be complete. Cautiously, add 2 mL of water to the flask to decompose any excess acetic anhydride. There will be some hot acetic acid vapor evolved as a result of the decomposition. When the liquid has stopped giving off vapors, remove the flask from the water bath and add 20 mL of water. Let the flask cool for a few minutes in air, during which time crystals of aspirin should begin to form. Put the flask in an ice bath to hasten crystallization and increase the yield of product. If crystals are slow to appear, it may be helpful to scratch the inside of the flask with a stirring rod. Leave the flask in the ice bath for at least 5 minutes. Collect the aspirin by filtering the cold liquid through a Buchner funnel using suction. Turn off the suction and pour about 5 mL of ice-cold distilled water over the crystals; after about 15 seconds turn on the suction to remove the wash liquid along with most of the impurities. Repeat the washing process with another 5-mL sample of ice-cold water. Draw air through the funnel for a few minutes to help dry the crystals and then transfer them to a piece of dry, weighed filter paper. Weigh the sample on the paper to ±0.1 g. Test the solubility properties of the aspirin by taking samples of the solid the size of a pea on your spatula and putting them in separate 1-mL samples of each of the following solvents and stirring: 1. toluene, C6H5CH3, nonpolar aromatic 2. heptane, C7H16, nonpolar aliphatic 3. ethyl acetate, C2H5OCOCH3, aliphatic ester Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Experiment 41

Preparation of Aspirin

327

4. ethyl alcohol, C2H5OH, polar aliphatic, hydrogen bonding 5. acetone, CH3COCH3, polar aliphatic, nonhydrogen bonding 6. water, highly polar, hydrogen bonding To determine the melting point of the aspirin, add a small amount of your prepared sample to a melting point tube (made from 5-mm tubing), as directed by your instructor. Shake the solid down by tapping the tube on the bench top, using enough solid to give you a depth of about 5 mm. Set up the apparatus shown in Figure 41.1. Fasten the melting point tube to the thermometer with a small rubber band, which should be above the surface of the oil. The thermometer bulb and sample should be about 2 cm above the bottom of the tube. Heat the oil bath gently, especially after the temperature gets above 100°C. As the melting point is approached, the crystals will begin to soften. Report the melting point as the temperature at which the last crystals disappear. To analyze your aspirin for its salicylic acid impurity, weigh out 0.10 ± 0.01 g of your sample into a weighed 100-mL beaker. Dissolve the solid in 5 mL 95% ethanol. Add 5 mL 0.025 M Fe(NO3)3 in 0.5 M HCl and 40 mL distilled water. Make all volume measurements with a graduated cylinder. Stir the solution to mix all reagents. Rinse out a spectrophotometer tube with a few milliliters of the solution and then fill the tube with that solution. Measure the absorbance of the solution at 525 nm. From the calibration curve or equation provided, calculate the percentage of salicylic acid in the aspirin sample. The contents of the suction flask may be poured down the drain. The toluene and heptane from the solubility tests should be put in the waste crock.


Figure 41.1



Name ____________________________________

Section ________________________________

Experiment 41

Data and Calculations: Preparation of Aspirin Mass of salicylic acid used

____________ g

Volume of acetic anhydride used

____________ mL

Mass of acetic anhydride used (density = 1.08 g/mL)

____________ g

Mass of aspirin obtained

____________ g

Theoretical yield of aspirin

____________ g

Percentage yield of aspirin

____________ %

Melting point of aspirin

____________ °C

Absorbance of aspirin solution

____________

Percentage of salicylic acid impurity

____________ %

Solubility properties of aspirin Toluene

____________

Ethyl alcohol

____________

Heptane

____________

Acetone

____________

Water

____________

Ethyl acetate ____________ S = soluble

I = insoluble

SS = slightly soluble

Underline those characteristics listed which would be likely to be present in a good solvent for aspirin. organic

aliphatic

polar

hydrogen bonding

inorganic

aromatic

nonpolar

nonhydrogen bonding



Name ____________________________________

Section ________________________________

Experiment 41

Advance Study Assignment: Preparation of Aspirin 1. Calculate the theoretical yield of aspirin to be obtained in this experiment, starting with 2.0 g of salicylic acid and 5.0 mL of acetic anhydride (density = 1.08 g/mL).

____________ g 2. If 1.6 g of aspirin were obtained in this experiment, what would be the percentage yield?

____________ % 3. The name acetic anhydride implies that the compound will react with water to form acetic acid. Write the equation for the reaction.

4. Identify R and R′ in Equation 1 when the ester, aspirin, is made from salicylic acid and acetic acid.

5. Write the equation for the reaction by which aspirin decomposes in an aqueous ethanol solution.



Experiment 42 Rate Studies on the Decomposition of Aspirin

nce aspirin has been prepared, it is relatively stable and in tablet form in a tight container resists decomposition for years. If it is exposed to moist air or heat, it will slowly break down by the reverse of the reaction we used to make aspirin in Experiment 41:

O

Aspirin + Water → Salicylic Acid + Acetic Acid

(1)

In this experiment we will study the rate at which aspirin in water solution will decompose at about 80°C. Like many organic reactions, this one is slow at room temperature. Raising the temperature allows us to carry out the reaction in a few minutes rather than several hours. To analyze the reaction mixture we will use the same method as in Experiment 41, where we find the amount of salicylic acid impurity in the aspirin sample. Iron(III) forms a very intensely colored violet complex with salicylic acid, so we can use the amount of light the reaction mixture absorbs to find out how far the reaction has gone. This will allow us to determine the rate of reaction at different concentrations. Our goal will be to use the observed rates to establish the order of the reaction, the rate constant, and the activation energy. The theory and terminology we will use are found in Experiment 20, which you should review now, even though you may have performed the experiment earlier in this course.

Experimental Procedure Obtain a digital thermometer from the stockroom.

A. Measuring the Rate of Reaction Step 1. Setting the stage. Set up a hot-water bath, using a 600-mL beaker about 3/4 full of water.

As a heater, use an electric hot plate with magnetic stirrer if such is available. Otherwise, use a Bunsen burner, supporting the beaker on a wire screen on an iron ring. If you use a burner, you should put a larger iron ring around the beaker, near the top, to prevent its tipping over. Clamp the ring to the ring stand. Prepare an ice bath by filling a 400-mL beaker with crushed ice; add water until the level is at the top of the ice. Clean three regular (18 × 150-mm) test tubes. Rinse them with a few mL of acetone; pour the acetone into one of the tubes, shake it, then pour it into the next, and the next. Dry the tubes in a gentle stream of compressed air. Number the test tubes and put them in a test tube rack. Put a small plastic funnel in Tube #1. Step 2. Weighing the sample. While the water is heating, weigh out on an analytical balance about

60 mg of aspirin, to 0.0001 g. This takes a bit of doing, since the sample is so small. It is probably easiest to make the weighing on a piece of weighing paper. Tare the paper, and use a small spatula to add a very small amount of aspirin. Do this several times, until you are within about 5 mg of the desired mass. Record the mass. Then, carefully, remove the paper from the balance. Fold it at the edge, and transfer the sample to the funnel. Tap the paper and the funnel to get as much sample as possible into Tube #1. Step 3. Preparing the reaction mixture. With a rinsed 10-mL graduated cylinder, or a pipet, add

10.0 mL of distilled water to the aspirin in Tube #1. The aspirin will not dissolve in the water; some will stay at the bottom of the tube, and some will float on the water. Put your stirring rod in the tube.


334

Experiment 42

Rate Studies on the Decomposition of Aspirin

Step 4. Carrying out the reaction. By now the water in the 600-mL beaker should be getting

warm. If it is boiling, slowly add tap water or ice to lower the temperature to 80°C, as measured to 0.1°C on your digital thermometer. C A U T I O N : Boiling water can give you a bad burn, so be careful not to tip over the beaker. Adjust your heat source as necessary to hold the bath steady at about 80°C, for at least five minutes. When you have managed that, and it may take a little time, you are ready to make your first run. Pick up the tube, and, noting the time to 1 second, put the tube in the water bath. Stir the mixture continuously, to help dissolve the aspirin and bring the reaction mixture up to temperature quickly. When the aspirin is all dissolved, which should happen within a minute, keep the rod in the tube, stirring occasionally. Measure and record the temperature of the bath at one-minute intervals as the reaction proceeds; it will fall initially, but should come back to the original temperature fairly quickly. Leave the tube in the bath for exactly six minutes. Then remove the tube and put it in the ice-water bath to stop the reaction. Stir the reaction mixture to hasten the cooling. Within a minute the solution should be cool, and you can put the tube back in the test tube rack. Step 5. Adding the indicator. Drain the graduated cylinder or pipet, and use it to measure out

5.0 mL of 0.025 M Fe(NO3)3 into the test tube. The solution should turn violet as the ironsalicylic acid complex forms. Stir the mixture until it is of uniform color. Step 6. Finding the salicylic acid concentration. Measure the absorbance of the solution at 525 nm

on a Spectronic 20 or other spectrophotometer. Use the procedure described in Appendix IV if this is the first time you have measured absorbance. After adjusting the spectrophotometer for 0 and 100% transmittance, rinse the spectrophotometer tube twice with a small amount of the solution from Step 5. Then add a few mL of that solution to the tube and record the absorbance you obtain. Find the concentration of the salicylic acid in the reaction mixture, using the graph that is provided or the equation you are given relating absorbance and concentration. This will be the salicylic acid concentration in the reaction mixture before the indicator was added. By Equation 1, that concentration will equal the change that occurs in the aspirin concentration. Step 7. Correcting for warmup time. Once the reaction mixture is up to 80°C, the reaction rate is

essentially constant, but during the first minute, when the aspirin is dissolving and the mixture is warming up, the reaction is much slower. To correct for this, you might run the same size sample at 80°C, for one minute, and subtract the absorbance for that sample from the total absorbance. You could then use 5 minutes as the time the reaction occurred. The authors actually did this, but found that if they took account of the warmup period by using the total absorbance and 5.5 minutes for the time of reaction, the rates were essentially the same as with the more accurate, but more complex, procedure. So, when calculating the rate, we will adopt the simpler approach and use 5.5 minutes as the time, and the total absorbance observed in Step 6.

B. Determining the Order of the Reaction and the Rate Constant Reset your water bath to the temperature you used in Part A, and prepare a new ice bath. Repeat Steps 1 through 7, at 80°C, using about 40 mg of sample, weighed to 0.0001 g, in Test Tube #2. Be sure to rinse out your graduated cylinder or pipet with distilled water before measuring out the 10 mL of water. The results you obtain with this sample, along with those from Part A will allow you to find the order of the reaction and the rate constant at 80°C. On the Data page carry out the calculations for the order and the rate constant of the reaction. Calculate the mean value of the rate constant and the standard deviation. (See Appendix VIII.) Then proceed to Part C.


Experiment 42

Rate Studies on the Decomposition of Aspirin

335

C. Finding the Activation Energy The last piece of information to be obtained in this experiment is the activation energy for the reaction. That is the energy that must be present in the key step when the reaction proceeds. It is found by measuring the dependence on temperature of the rate constant for the reaction. We will use nearly the same procedure as in Part A, Steps 1 through 7, except that we will carry out the reaction in Test Tube #3 at about 70°C instead of 80°C. The sample size should be about 60 mg, weighed accurately. Bring the water bath to about 70°C, or a bit higher, and hold it there for five minutes. Make a new ice bath, and make sure that you rinse your pipet or graduated cylinder before measuring out the water. At 70°C the reaction will go a lot more slowly than in Part A. In order to get easily measured absorbance values, run the reaction for twelve minutes, rather than six. The aspirin will be slower to dissolve, so the warmup period will be two minutes, rather than one. It will require a considerably longer time to dissolve the aspirin, but, with continuous stirring, it should go into solution in two minutes. Record the temperature at 2-minute intervals. Record the absorbance that you obtain. To correct for warmup time, calculate the reaction rate using the total absorbance and time equal to 11 minutes. With the data you have obtained, you can proceed to evaluate the energy of activation, Ea, for the reaction. This is done using the rate constants at 80°C and 70°C and the Arrhenius equation, as in Experiment 20: ln k = −Ea/RT + a constant DISPOSAL OF REACTION PRODUCTS.

(2)

All of the reaction products may be poured down the sink,

with the water running as you pour.



Name ____________________________________

Section ________________________________

Experiment 42

Data and Calculations: Rate Studies on the Decomposition of Aspirin Let Asp = aspirin and Sal = salicylic acid (MM Asp = 180 g) Part A Part B Part C Tube #1 Tube #2 Tube #3 ____________ ____________ ____________

1. Mass of Asp in tube in g No. moles Asp in tube

____________ ____________ ____________

[Asp] in moles/L (10 mL)

____________ ____________ ____________

(Temperature measurements are on following page) 2. Absorbance, Abs, of solution

____________ ____________ ____________

[Sal] in reaction mixture

__________ M __________ M __________ M

Change in [Asp], Δ[Asp] = −[Sal]:

__________ M __________ M __________ M

Note that Δ[Asp]

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